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TEXT BOOK 



OF 



Elementary Mechanics, 



FOR THE USE OF 



COLLEGES AND SCHOOLS. 



BY 

EDWARD S. DANA, 

Assistant Professor of Natural Philosophy in Yale College, 



ELEVENTH EDITION 



NEW YORK: 

JOHN WILEY AND SONS, 

53 East Tenth Street 

Second door west of Broadway. 
1890. 



QAsot 

0)17 



Copyright, 1881, 
By EDWARD S. DANA. 






PREFACE 



The writer has been induced to prepare this Text 
Book because of his inability to find, among the many 
excellent works on Mechanics, one that was thoroughly 
adapted to his special wants, and because it has seemed 
to him probable that similar needs must haye been felt 
by other instructors. 

The chief aim has been to present the fundamental 
principles of the subject in logical order, and in as clear, 
simple, and concise a form as possible, yet without any 
sacrifice of strict accuracy. For the sake of making 
the portions of the subject, which necessarily involve 
some difficulty, more intelligible to beginners, and also 
to increase the interest of the general principles demon- 
strated by showing something of their practical bearings, 
simple illustrations have been introduced rather more 
fully than usual; these are sometimes given in a few 
words, sometimes in more extended form. This has 
led to a slight expansion of the book in size, but does not 
proportionately increase the time required to master it. 
The general scheme is not more extended than the sub- 
ject demands, and, if time is limited, the instructor 
can readily select those articles whose omission will not 
interfere with the completeness of the study. [Some 



IV PREFACE. 

of the articles which, if necessary, may be omitted, are: 
26, 28, 36, 43, 45, 50, 51, 70, 75, 76, 81, 100, 105, 106, 
129, 131 a, I, c, 135, 139, 142, 146, 149, 155, 158, 171, 
177, 189, 190, 193, 196, 197, 198, 199, 206, 208, 211, 213, 
214, 215, 216, 224, 225, 230, 231, 235, 239, 241, 242, 
243, 250, 251. To this list, which might be somewhat 
extended, can be added the matter in fine print.] 

The study of Elementary Mechanics is one of very 
great value in a course of liberal education. The sub- 
ject, when properly presented, affords an excellent kind 
of mathematical training, which is suitable to all grades 
of students ; it furnishes applications of the principles 
and methods previously learned in Geometry and Trigo- 
nometry, and further has the advantage of dealing with 
the real phenomena of nature, and is not confined to 
abstract principles; it is also a necessary introduction to 
Physics. With reference to the last end, the chapter 
on Work and Energy has been expanded to considerable 
length, but not greater than the importance of those 
principles justifies. 

The book is limited to the Mechanics of Solids, be- 
cause that forms a complete subject by itself, while an 
equally extended discussion of Liquids and Gases would 
be obviously out of the question. Moreover, the latter 
subjects are treated at length from their experimental 
side in Text Books on Physics, and that is about all that 
the general student requires. 

Examples are given at the end of each division of tha 
subject, designed to show the most important applica 
tions of the principles; answers to them will be foundf 
on pages 279-286. By simply changing the numbers in 
these examples they may be increased indefinitely with- 
out adding seriously to the work of the instructor. The 



PEEFACE. V 

examples are, in general, arranged so as to involve but 
little mechanical labor of computation — long calcula- 
tions tending rather to obscure than make clear the 
principles for the illustration of which the examples are 
given. As it is desirable that every student should have 
some knowledge of the common metric units, a series of 
examples introducing them is added at the close of the 
volume. 

The author takes pleasure in acknowledging his in- 
debtedness to Prof. H. A. Newton, who has rendered 
him very important aid while the book was passing 
through the press, and also to Dr. J. J. Skinner, of the 
Sheffield Scientific School, for numerous valuable sug- 
gestions. 

New Haven, Conn., January 1, 1881. 



TABLE OF CONTENTS. 



INTRODUCTION. 



Matter. 2. Body; particle. 3. Molecule. 4. Physical science. 
5. Atom. 6. Chemistry. 7. States of matter: solid, liquid, 
gaseous. 8, 9. Properties of matter. 10. Mechanics: Kine- 
matics, Dynamics (or Kinetics), Statics Pages 1-5 

Brief explanation of the metric system Pages 5, 6 



CHAPTER I— KINEMATICS. 

Motion and Rest — Kinds of Motion. 

11. Motion and rest. 12. Kinds of motion. 13. Motion of trans- 
lation. 14. Motion of rotation. 15. Translation and rotation. 
16. Path of a particle or body Pages 7-9 

Uniform Motion. 

17. Uniform motion; constant velocity. 18. Constant angular 
velocity. 19. Space passed over in uniform motion. 20. Geo- 
metrical representation of velocity. 21. Geometrical represen- 
tation of the space passed over in uniform motion. .Pages 9-12 
I. Examples : Uniform motion of translation or rota- 
tion Pages 12, 13 

Varied Motion — Acceleration. 

22. Varied motion; variable velocity. 23. Acceleration. 24. 
Velocity acquired in uniformly accelerated motion. 25. Space 
passed over in uniformly accelerated motion. 26. Average 
velocity. 27. 28. Formulas for accelerated motion. .Pages 13-20 



Vlll CONTENTS. 

II. Examples : Uniformly accelerated motion ; Falling 
bodies, etc Pages 20-22 

Composition and Resolution of Velocities — Uniform Motion. 

29. Composition of motions in general. 30. Resultant and com- 
ponent velocities. 31. Composition of constant velocities in 
the same straight line. 32. Composition of two constant 
velocities not in the same straight line. 33. Parallelogram 
op Velocities. 34, 35. Calculation of the magnitude and 
direction of the resultant velocity. 36. Illustrations. 37. 
Composition of several constant velocities. 38. Resolution of 

velocities Pages 22-29 

III., IY. Examples: Composition and resolution of con- 
stant velocities Pages 30-32 

Composition and Resolution of Accelerations. 

39. Composition and resolution of accelerations. 40. Motion 

down an inclined plane Pages 32, 33 

V. Examples : Bodies falling down an inclined plane. 

Pages 33, 34 

Composition of Uniform and Accelerated Motion in the same 
Straight Line. 

41, 42. Composition of uniform and accelerated motion in the 
same straight line. 43. Geometrical representation. 44. 
Motion of a body projected vertically upward. 45. Projected 

up or down an inclined plane Pages 34-39 

VI., VII., VIII., IX. Examples: Bodies projected vertically 
downward; bodies projected vertically upward; bodies 
projected up or down a smooth inclined plane; bodies 
projected against friction Pages 39-42 

Composition of Uniform and Accelerated Motion not in the same 
Straight Line. 

46. Composition of uniform and accelerated motion. 47, 48, 49, 

50, 51. Projectile Pages 42-50 

X. Examples: Projectiles Page 50 



CONTENTS. IX 



CHAPTER II.— DYNAMICS, OR KINETICS. 

Mass — Density — Volume — Momentum. 

52. Dynamics, or Kinetics. 53. Mass or quantity of matter. 
54. Mass determined by weight. 55. Distinction between 
mass and weight. 56. Relation between mass, density, and 
Yolume. 57. Momentum Pages 51-55 

XI. Examples: Mass; density; volume Page 55 

58. Definition of Force. 59. Continued and impulsive forces. 

60. Effects of force upon a free body. 61. Equilibrium. 62. 
Examples of force. 63, 64, 65. Force of gravity. ..Pages 55-62 

XII. Examples : Force of gravity Page 63 

Newton's Laws of Motion. 

66. Laws op Motion. 67. First law of motion explained. 63. 
Second law of motion explained; deduction of dynamical 
formulas. 69. Third law of motion explained. 70. Collision 
of inelastic bodies Pages 63-71 

XIII. Examples: Collision of inelastic bodies Page 72 

Measurement of Force. 
71. Absolute method of measuring force. 72. Gravitation 
method of measuring force Pages 72-75 

Problems in Dynamics. 

73. Dynamical formulas. 74. Attwood's machine. 75, 76. Dy- 
namical problems Pages 75-80 

XIV. Examples: General dynamical problems.. Pages 80-82 



CHAPTER III.— DYNAMICS— CENTRAL FORCES. 

77. Uniform circular motion ; centrifugal and centripetal forces. 
78. Calculation of the acceleration of a central force. 79. 
Illustrations of centrifugal force. 80. Centrifugal force due 

to the earth's rotation Pages 83-88 

XV. Examples: Centripetal and centrifugal forces. 

Pages 88, 89 



CONTENTS. 



CHAPTER IV.— DYNAMICS— FRICTION. 

I. Definition of friction. 83. Reaction of smooth surfaces. 
84. Lubricators, etc. 85. Effects of friction. 86. Kinds of 
friction : sliding and rolling. 87. Fluid friction. 88. Laws of 
friction. 89, 90. Explanation of laws of friction. 91. Co- 
efficient of friction. 92. Angle of friction. 93. Determination 
of the coefficient of friction. 94. Examples of the coefficient 

of friction Pages 90-99 

XVI. Examples: Friction Pages 99, 100 



CHAPTER V.— DYNAMICS— WORK AND ENERGY. 

A. Mechanical Work — Measurement op Work. 

95. Definition of work. 96. Examples of work. 97, 98. Meas- 
urement of work. 99. Rate of work. 100. Application of 
principles of work Pages 101-105 

XVII. Examples: Work Pages 105, 106 

B. Energy— Conservation and Correlation of Energy. 

101. Definition of energy. 102. Forms of energy. 103. Kinetic 
and potential energy. 104, 105. Measurement of energy. 
106. Relation of kinetic energy to momentum. 107. Trans- 
formation of kinetic and potential energy. 108. Apparent loss 
of visible energy. 109. Nature of heat. 110. Examples of 
the production of heat from mechanical energy. 111. Definite 
relation between heat and mechanical work. 112. Conversion 
of heat into work. 113. Other forms of molecular energy. 
114. Examples of the transformation of energy. 115. Conser- 
vation of energy. 116. Terrestrial stores of energy. 117. 
The sun as the ultimate source of terrestrial energy. 118. 
Dissipation of energy Pages 106-124 

XVIII. Examples: Potential and kinetic energy. 

Pages 124^126 



CONTENTS. XI 



CHAPTER VI.— STATICS. 

Introductory. 

119. Statics. 120. Geometrical representation of a fore*,. 121. 
Line of action of a force. 122. Transmission of a force in its 
line of action. 123. Body; particle Pages 127, 128 

Composition of Forces meeting in a Point. 

124 Composition of forces; resultant and component forces. 
125. General condition of equilibrium. 126. Composition of 
forces having the same line of action. 127. Condition of 
equilibrium for forces having the same line of action. 128. 
Parallelogram of Forces. 129. Experimental verification 
of the parallelogram of forces. 130, 131. Calculation of the 
resultant. 132, 133. Conditions of equilibrium for three forces 
acting on a particle. 134. Composition of more than two forces 
acting in the same plane upon a particle. 135. Forces not in 
the same plane. 136. Condition of equilibrium for more than 
three forces acting on a particle Pages 129-142 

XIX. Examples: Parallelogram of forces Pages 142, 143 

Resolution of Forces. 

137. Resolution of forces in general. 138. Rectangular com- 
ponents. 139. Illustration of the resolution of forces. 140. 
Resolution of forces along two axes at right angles to each 
other. 141. Condition of equilibrium for three or more forces 
acting on a particle. 142. Resolution of forces along three 
axes Pages 143-150 

XX, XXI. Examples: Resolution of forces. .Pages 150-152 

Composition and Resolution of Parallel Forces. 

143. Parallel forces. 144. Like parallel forces. 145. Unlike 
parallel forces. 146. Experimental verification. 147. Three 
or more parallel forces. 148. Three parallel forces in equi- 
librium. 149. Resolution of parallel forces. 150. Couples. 

Pages 152-159 
XXII. Examples: Parallel forces Pages 159, 160 



Xll CONTENTS. 

Forces tending to produce Rotation — Moments. 
151, 152. Moment. 153. Positive and negative moments. 154. 
Geometrical representation of the moment of a force. 155. 
Proposition in regard to moments. 156. Equality of moments; 
principle of the lever. 157. Free and constrained body. 

Pages 161-167 

XXIII. Examples: Moments Page 167 

Summary of Conditions of Equilibrium. 

158. Summary of conditions of equilibrium for forces acting on 
a body in one plane Pages 167, 168 

CHAPTER VII.— STATICS— CENTRE OF GRAVITY. 
A. Centre op Gravity op Bodies — Plane and Solid. 

159. Definition of the centre of gravity. 160. Centre of gravity 
of two bodies. 161. Centre of gravity of any number of 
bodies. 162. Centre of gravity of a straight line. 163. Centre 
of gravity of a plane figure determined by its symmetry. 164. 
Centre of gravity of regular polygons. 165. Centre of gravity 
of a parallelogram. 166. Centre of gravity of a triangle. 167. 
Centre of gravity of a solid figure. 168. Centre of gravity of 
a triangular pyramid. 169. Centre of gravity of any pyramid. 
170. Centre of gravity of a cone. 171. Problems. .Pages 169-180 

XXIV. Examples: Centre of gravity Pages 180-182 

B. Application op the Principles of the Centre op 
Gravity— Equilibrium and Stability. 
172. Conditions of equilibrium. 173. Experimental determina- 
tion of the centre of gravity. 174, 175. Stable, unstable, and 
neutral equilibrium. 176. Stability of a body resting on a 
base. 177. Conditions upon which the stability of a body 
depends Pages 182-189 

XXV. Examples: Stability Pages 189, 190 

CHAPTER VIII— STATICS— MACHINES. 

178. 179, 180, 181. Principle of work as applied to the machines. 
182. Virtual velocities. 183. The machines with friction. 
184. Simple machines Pages 191-195 



CONTENTS. Xlll 

I. Lever. 

A. General Principles of the Lever. 

185. The lever defined. 186. Relation of the power and weight 
in the lever. 187. Three kinds of lever. 188, 189. Illustration 
of the lever. 190. The lever on the principle of work 

Pages 195-201 

B. Some Special Applications of tlie Principle of the Lever. 

I. BALANCE. 

191. Balance. 192, 193. Conditions to be fulfilled by a good 
balance Pages 202-205 

H. STEELYARD. 

194. Common steelyard. 195. Application of the steelyard. 
196. Danish steelyard. 197. Roberval's balance.. Pages 205-210 

m. TOGGLE-JOINT. 

198. Relation of power to weight in the toggle-joint. 199. Stone- 
crushing machine Pages 210-212 

TV. COMPOUND LEVERS. 

200. Relation of power to weight in the compound levers. 201. 

Application of compound levers Pages 212, 213 

XXVI., XXVII. , XXVIII. Examples: Lever; balance; 
steelyard Pages 213-215 

II. Wheel and Axle. 

202. Wheel and axle. 203, 204, 205. Relation of power to weight 
in wheel and axle. 206. Wheel and axle on the principle of 
work. 207. Applications of the wheel and axle. 208. Chinese 

windlass Pages 215-219 

XXIX Examples: Wheel and axle Page 220 

III. Toothed Wheels. 

209. Toothed wheels. 210. Relation of power to weight with 
the toothed wheels. 211. Toothed wheels on the principle of 
work. 212, 213, 214, 215. Applications of the toothed wheels. 
216. Use of belts Pages 220-226 



XIV CONTENTS. 

IV. Pulley. 
217. Pulley. 218. Single fixed pulley. 219. Single movable 
pulley with parallel strings. 220. SiDgle movable pulley with 
inclined strings. 221, 222, 223, 224. Combinations of pulleys. 
225. The pulley on the principle of work. 226. Application 
of the pulley Pages 226-234 

XXX. Examples: Pulley Pages 234, 235 

V. Inclined Plane. 

227. Inclined plane. 228, 229, 230. Relation of the power and 

weight for the inclined plane. 231. The inclined plane on 

the principle of work. 232. Applications of the inclined 

plane Pages 235-241 

XXXI. Examples: Inclined plane Pages 242, 243 

VI. Wedge. 

233. Wedge. 234. Relation of the power to the weight for the 
wedge. 235. The wedge ou the principle of work. 236. Appli- 
cation of the wedge Pages 243-245 

XXXII. Examples: Wedge Page 245 

VII. Screw. 

237. Screw. 238. Relation of the power to the weight for the 
screw. 239. The screw on the principle of w T ork. 240. Appli- 
cation of the screw. 241. Micrometer screw. 242. Differen- 
tial screw. 243. Endless screw Pages 246-251 

XXXIII. Examples: Screw Page 251 

CHAPTER IX.— PENDULUM. 

244. Motion in a vertical circle. 245. Motion of a simple pendu- 
lum. 246, 247. Compound pendulum. 248. Application of 
the pendulum. 249, 250. Values of I and g. 251. Other appli- 
cations of the pendulum Pages 252-261 

XXXIV. Examples: Pendulum Page 261 

Additional Examples, introducing the Metric Units. 

Pages 263-278 
Answers to Examples Pages 279-291 



ELEMENTAEY MECHANICS. 



INTRODUCTION". 



1. Matter. Matter is the substance of which bodies 
are composed; it is that which maybe apprehended by 
the senses, and which may be acted upon by force. 

2. Body-Particle. A lody is any portion of matter 
which is bounded in every direction. A material parti- 
cle is a body of dimensions so small that it is unnecessary 
to consider the differences in position or motion of its 
different parts. 

In many cases the differences in the relations of the 
parts of an extended body are, in like manner, left out 
of account, it being considered as a single unit, and then 
the body is treated as a particle. 

3. Molecule. The smallest portion into which a 
given kind of matter can be conceived to be divided, 
without a loss of its properties, is called the molecule. 
The molecule is an ideal unit, the existence of which is 
believed to be proved by experiment, although it cannot 
be by direct observation. The smallest portion of 
matter, obtained by any method of mechanical subdivi- 
sion, would consist of a large number of molecules. 

According to the conclusions of Sir William Thomson, 
if a drop of water were to be magnified to the size of the 
earth, the molecules, of which it is made up, would be 
coarser than fine shot and probably finer than cricket- 
balls. 



2 INTEODUCTIOlSr. [4. 

4. Physical Science. All changes which involve a 
material body, either as a whole or with respect to the 
relations of its molecules, are considered under the head 
of Natural Philosophy, or Physical Science. Thus, the 
fall of a body to the earth; the flight of a rifle-ball; the 
ring of a bell; the melting of iron, and its contraction 
or expansion on change of temperature, its magnetiza- 
tion — these and all other analogous phenomena are in- 
cluded under Physical Science. 

5. Atom. Every molecule is supposed to be made up 
of one or more indivisible units called atoms ( a priv. 
and repLVGD, to divide). Thus, the smallest conceivable 
particle, or molecule, of salt, possessing all the proper- 
ties of the mass, is believed to consist of two dissimilar 
atoms, one of the metal sodium, the other of the gas 
chlorine. 

6. Chemistry. All phenomena which result in a re- 
arrangement of the atoms and a consequent change in 
the molecules of a body — that is, a loss of identity of the 
substance involved — belong to Chemistry. For example, 
the change of ice to water, or of water to steam, involves 
no change in the molecules but only in their mutual re- 
lations and position, hence these phenomena belong to 
Physical Science; but when a rearrangement of the atoms 
takes place and the water is thus decomposed into its 
constituent gases, hydrogen and oxygen, this last is a 
chemical change. 

The molecule is the physical unit; the atom is the 
chemical unit. 

7. States of Matter. Matter may exist in three differ- 
ent states : the solid, liquid, and gaseous states. 

The solid is characterized by a greater or less degree 



9.] INTRODUCTION. 3 

of rigidity. The molecules are bound together by the 
molecular force of attraction, called cohesion, and hence 
a solid body tends to retain its own shape. 

The liquid is characterized by its mobility; the mole- 
cules are free to move about each other, and the liquid 
takes the shape of any containing vessel. 

The gas is characterized by its tendency to indefinite 
expansion. The molecules are believed to be in rapid 
motion and constantly coming into collision and then 
repelling one another, so that a gas tends to occupy a 
greater volume, and hence exerts pressure on the sides 
of any vessel in which it is confined. 

The term fluid is sometimes employed to include both 
liquids and gases. 

Many substances may under varying conditions exist 
in the three different states : this is illustrated by the 
case of water, which is a solid — ice — below the freezing 
point, a liquid at ordinary temperatures, and a gas — steam 
• — at high temperatures. 

8. Properties of Matter. All forms of matter possess 
the essential properties of extension, impenetrability, and 
inertia. 

(1) Extension : Every body occupies a definite por- 
tion of space; that is, it has length, breadth, and thick- 
ness. 

(2) Impenetrability : Two forms of matter cannot 
occupy the same space at the same time. 

(3) Inertia : Matter has no power to change its 
own state of motion or rest, hence it offers an apparent 
resistance to a force tending to change its state. This is 
further explained in a subsequent article (67). 

9. Other properties of matter are — 

Porosity : The molecules, of which a given body is 



4 INTEODUCTION, [10. 

supposed to be made up, are believed to be separated 
from one another by a greater or less space. In addi- 
tion to these true or physical pores, most bodies exhibit 
also visible open spaces, or sensible pores, as those of a 
sponge. 

Compkessibilitt : A body may be made by pressure 
to occupy a smaller space ; this is a direct consequence 
of its porosity. 

Divisibility : A given kind of matter admits of be- 
ing divided into a very great number of parts. 

Elasticity : A body, whose shape has been altered by 
a force acting on it, tends to regain its shape when the 
force ceases to act. Solids vary widely in elasticity : for 
example, compare lead and steel, or clay and ivory. 
Liquids and gases are perfectly elastic. 



10. Mechanics is that branch of Physical Science 
which considers the motion and equilibrium of bodies. 
Corresponding to the three states of matter, the subject 
of Mechanics is divided into 

(1) Mechanics of solids. 

(2) Mechanics of liquids, including Hydrostatics and 
Hydrodynamics. 

(3) Mechanics of gases, or Pneumatics. 

The first of these three divisions, which forms the 
subject of this text-book, is further divided into three 
parts, Kinematics, Dynamics or Kinetics, and Statics. 

Kinematics * includes the discussion of abstract 
motion; that is, of the motion of bodies without refer- 
ence to their mass (quantity of matter), or to the force or 

* From the Greek nivrjfia, motion. 



10.] INTRODUCTION". 5 

forces which cause their motion. To the idea of space, 
involved in Geometry, it adds that of time. 

Dynamics,* or Kinetics,! embraces the discussion of 
the action of a force, or of forces, in producing the mo- 
tion of bodies of known mass. 

Statics| discusses the action of forces upon bodies in 
so far as they hold the body acted upon at rest; that is, 
in equilibrium. 

*From the Greek SvvajuiS, power. 

f From Kir ego, to move, 

X From drarixoS (i6z7jfit), causing to stand. 

METRIC SYSTEM. 

UNITS OF LENGTH. 

ENGLISH UNITS. , 

Kilometer 1000 meters. 3280.9 feet. .62137 mile. 

Meter.... 39.37 inches. 3.281 feet. 

Decimeter 1 meter. 3.937 " .3281 " 

Centimeter 01 " 0.3937 " .0328 " 

Millimeter {mm).. .001 " 0.0394 " .00328 " 

UNITS OF VOLUME. 

DRY MEASURE. LIQUID MEASURE. 

Kiloliter (or Stere) 1 cub. meter, 1.308 cub. yds. 264.17 galls. 

Liter 1 cub. decimeter, 0.908 quarts, 1.0567 quarts. 

Milliliter 1 cub. centimeter, 0.061 cub. in. 0.27 fl. dr'hm. 

UNITS OF WEIGHT. 
VOLUME OP WATER AVOIRDUPOIS MEASURE. 

giving the weight. (1 lb. =7000 grains.) 

Kilogram cub. dec'm'r or liter. 15432.3 grains. 2.2046 lbs. 

Gram cub. centimeter. 15.432 " 0.0022 " 

Milligram .... cub. millimeter. 0. 0154' ■ 0. 0000022" 



METEIC SYSTEM. 

•& Meter = 1 Decimeter = 10 Centimeters = 100 Millimeters. 



.Centimeters." 
5 4 



TTTlfTTT TtTTTTtTTTTtTT 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Mill 111 III ll II I ill 1 1 1 ll I lllllll ll 1 1 1 ll 1 1 1 llll ll 1 1 II ll 1 1 1 1 1 Ml 



~^^J l I i | * M j M * | * I ■» | M » | 



I I I I l I I l"T 



2 
Inchks. 



4 inches, each divided into eighths. 

The Meier is the length at 0° C (temperature of melting ice) of a 
certain platinum bar kept at Paris. It was intended to be (and is 
very nearly) equal to one forty-millionth part of the earth's cir- 
cumference about the poles. It is the only arbitrary unit of the 
Metric System, since all the other units of weight, etc. , are directly 
deiived from it. 

Units of Length. The meter is divided into 10 decimeters, 
into 100 centimeter*, into 1000 millimeters. Also, 10 meters^l deka- 
meier, .100 meters ~ 1 hectometer, 1000 meters =1 kilometer. (The 
same prefixes are used in a case of the other units, with a similar 
signification.) 

The approximate value of some of these units are as follows : — 
The meter is a little longer than the English yard; it is very nearly 
equal to 3 feet 3f inches. The millimeter is a little less than .04 of 
an inch, or 1 inch is a little more than 25 millimeters. (See figure 
above,) The kilometer is about £ of a mile. 

Units of Surface. The squares of the units of length are 
taken as the units of surface. The principal units are the centare, 
or square meter ; the are ( = 100 square meters) ; and the hectare 
(=10,000 square meters); the hectare is equal to 2.47 acres. 

Units of Volume. The cubes of the units of length are taken 
as the units of volume or capacity. The principal units are the 
cubic meter or stere, equal to 1.3 cubic yards; the cubic decimeter 
or liter, which is a little larger than a wine quart; and the cubic 
centimeter. 

Units of Weight. The weights of the units of volume of 
water (at 4°C = 39°. 2 F when it has its greatest density) are taken as 
the units of weight. The principal units of weight are the kilogram 
(or kilo), which is the weight of a liter or cubic decimeter of water 
at 4° C; it is equal to 2.2 pounds; and the gram, which is the 
weight of the cubic centimeter of water at 4° C; it is equal to about 
15 grains. 

The equivalents of the important metric units are given more 
exactly on the preceding page. 



CHAPTER I.— KINEMATICS. 

Motion and Rest — Kinds of Motion. 

11. Motion. A body is said to move when, in succes- 
sive intervals of time, it occupies different positions with, 
reference to some other body considered to be at rest. 

The terms motion and rest are simply relative, for the 
state of any body in this respect can be judged of only 
by comparing it with some other body or bodies. For 
example, the objects on the deck of a steamboat may be 
at rest with reference to each other and to the boat, while 
they are in motion as regards the neighboring shore. 
Again, two trains moving side by side at the same speed 
may seem to a passenger on either to be at rest, and are 
actually so as regards each other, while they are in rapid 
motion as regards the ground over which they are pass- 
ing. 

As the term rest is ordinarily employed in Mechanics, 
the earth is used as the basis of comparison, and in this 
sense bodies are said to be at rest which do not move 
with reference to it, as, for example, the buildings in a 
city. It is to be remembered, however, that the earth 
itself and hence all objects upon it are really moving 
very rapidly through space. In fact we know nothing 
of absolute rest, for all bodies of which we have any 
knowledge are in motion. 

Further than this, there is reason to believe that, in- 
dependent of the motion of the bodies themselves, the 



8 KINEMATICS. [12. 

molecules which make them up have also in all cases a 
very rapid vibratory motion of their own. Motion is 
then the actual state of matter so far as we know it, 
while the rest we observe is only apparent. 

12. Kinds of Motion. With respect to its direction, a 
body may have either motion of translation or of rota- 
tion. With respect to its rate, the motion may be uni- 
form or varied. 

13. Motion of Translation. If the motion of a body 
is such that every point in it has the same velocity, and 
every straight line in it remains parallel to itself, the 
body is said to have motion of translation. This is illus- 
trated by the motion of a sled down a hill, or that of the 
body of a carriage. 

The motion of translation of a particle may be either 
(1) rectilinear — that is, in a straight line — or (2) curvi- 
linear, in a curved line. 

14. Motion of Rotation. A body is said to have motion 
of rotation, or simply to rotate, when it moves about an 
axis so that the different particles describe concentric 
circles around it, their velocity increasing with their dis- 
tance from the axis. This is illustrated by the turning 
of a wheel on its axle, or the spinning of a top. 

15. A body may at the same time have both kinds of 
motion. For example, the wheel of a carriage rotates 
about its axle, and also moves forward — that is, has 
motion of translation — with the rest of the vehicle ; if 
the wheel is blocked, as in descending a steep hill, then 
it has motion of translation only. Again, the earth has 
a motion of rotation about its axis and also of transla- 
tion in its orbit about the sun. 

In the statements which follow in regard to the motion 



17.] UNIFORM MOTION. 9 

of bodies, motion of translation without rotation is 
always to be understood unless it is distinctly stated 
otherwise. 

As explained in Art. 2, the term body may be used 
instead of particle, when the body is considered as a 
unit, any distinction between the position or motion 
of the different parts being left out of accounf ; in this 
sense the term body is employed in the following arti- 
cles. 

16. Path of a Particle or Body. The path of a parti- 
cle, or trajectory as it is sometimes called, is the con- 
tinuous line, either straight or curved, which it describes 
as it moyes. By the path of a body is ordinarily meant 
the line described by some definite point in it, usually 
the centre of gravity, or the geometrical centre. 

Uniform Motion. 

17. Uniform Motion. The motion of a body is said to 
be uniform if it moves over equal spaces in equal suc- 
cessive intervals of time, however small these be taken. 
The velocity, or rate of motion, is then said to be con- 
stant. 

Constant velocity, or the velocity of a tody moving 
uniformly, is measured by the number of units of linear 
space passed over in the unit of time. 

The unit or space commonly employed is the foot, 
and of time the second. The unit of velocity is then 
a velocity of one foot per second ; this is a compound 
unit sometimes called a foot-second. Thus a constant 
velocity of 10 would mean that the body passed over 10 
feet in each successive second. 



10 KINEMATICS. [18. 

Other units of space and time are also not infre- 
quently employed : we speak of the velocity of the 
earth in its orbit as 19 miles per second ; of a train as 
so many miles an hour, and so on. If the metric system 
is made use of, the units of distance belonging to it must 
be taken ; that is, the millimeter, meter, kilometer, etc. 
18. Constant Angular Velocity. The statement in the 
preceding article has reference only to linear velocity. 
When, however, a body rotates on an axis and each 
particle describes a circle about it, it is often convenient 
to have an expression for the angular velocity. 

The angular velocity of a tody, 
rotating uniformly about an axis, 
is measured by the angle described 
in the unit of time by a radius 
moving in a plane perpendicular 
to the axis of rotation. This angle, 
as ACB (Fig. 1), is expressed not 
Fig. i. in degrees but in circular measure; 

that is, by the ratio of the arc to the radius [~Tp\ 

This angular velocity is usually represented by the letter 
go. Hence 

AB 

If AG = r, and AB = v = the linear velocity, then 

co = — , and . \ v = cor. 
r 

It is evident that the angular velocity is constant for all 
parts of a body rotating uniformly, but the linear velo- 
city increases directly with the distance from the axis. 




21.] OTIFOKM MOTIOK. 11 

19. Space passed over in Uniform Motion. If a body 
moyes uniformly for a time t, with a velocity v, the 
space, or distance (s), passed over is equal to the product 
of the time and velocity : 

s = vt; 

s s 

also. v = -7, and t = — . 

' t v 

20. Geometrical Representation of Velocity. The ve-* 
locity of a body may be represented geometrically by 
a straight line, whose direction is the direction of the 
motion, and whose length is taken proportional to the 
velocity. Thus (Fig. 2), if the motion of a particle be 





Fig. 2. Fig. 3. 

in the direction from A toward C with a velocity of 20, 
and in another independent case from A toward B with 
a velocity of 10, then these lines, if proportional to 20 
and 10 respectively, may be taken as representing these 
velocities geometrically ; here obviously AC = 2AB. 

If the particle moves in a curved path, as from M 
toward N (Fig. 3), its velocity at any points, as A and O, 
will be represented by tangents at these points, AB and 
CD, whose lengths are proportional to the velocities 
respectively. 

21. Geometrical Representation of the Space passed 
over in Uniform Motion. The space passed over by a 




12 KINEMATICS. [21. 

body moving uniformly for a given time may be repre- 
sented by the area of a rectangle, whose adjacent sides 
are taken proportional respectively 
to the time and velocity, each in 
terms of its own unit. Thus, sup- 
pose a body to move for t seconds 
with the constant velocity v; let 
AB (Fig. 4) be taken proportional 
to t, and BO to v. Then, since (19) 

s = vt, and area of rectangle = BC.AB, 

the space is proportional to, or, in other words, is repre- 
sented by, the rectangle. 

This principle, which is of interest chiefly from the 
part it takes in a subsequent demonstration (25), means 
simply that the relation of the area of the rectangle to 
its sides is the same as that of the space in uniform 
motion to the velocity and time. 

EXAMPLES. 
I. Uniform Motion of Translation or dotation. Articles 17-21. 

1. A body travels 30 feet per second: How far will it go in a 
day of 24 hours? 

2. A velocity of 30 miles per hour corresponds to a rate of how 
many feet per second? 

3. A man walks uniformly 4 miles per hour: (a) How many 
feet does he go in a second? (b) How many yards in a minute? 

4. Two bodies start from the same point in opposite directions, 
the one moves at a rate of 11 feet per second, the other at a rate 
of 15 miles per hour: (a) What will be the distance between them 
at the end of 8 minutes? (b) When will they be 825 feet apart? 

5. How far will the bodies in the preceding example be apart 
at the end of the same time, if they move in the same direction ? 

6. Two bodies, starting from the same point, move along lines 



22.] VARIED MOTION — ACCELERATION. 13 

at right angles to each other, the first at the rate of 4£ feet per 
second, the second at a rate of 200 yards per minute : How far 
will they be apart at the end of an hour? 

7. Suppose the earth travels in its orbit 600 million miles in 
365£days: What velocity has it, expressed in miles per second, 
supposing that the motion is uniform? 



8. "What is the linear velocity of a point on the equator due to 
the earth's rotation? — take the equatorial radius as 4000 miles. 

9. What is the linear velocity of a point on the earth at latitude 
60° from the same cause? 

10. If the linear velocity of a point at the equator, due to the 
earth's rotation, is ■», show that the velocity at any latitude (l) is 
equal to v cos I. 

11. What is the angular velocity of the earth's rotation per 
second? 

12. (a) What is the angular velocity of the fly-wheel of an 
engine, 6 feet in diameter, if it makes 40 revolutions in a minute? 
(b) What is the linear velocity of a point on the circumference? 

13. (a) What is the angular velocity of a buzz-saw, having a 
radius of 2 feet, if it makes 100 revolutions per second? (b) How 
far (in miles) will a point on the circumference travel in a work- 
ing day of 10 hours? 

14. The angular velocity of awheel is \it per second: What is. 
the linear velocity of points at distances of (a) 2 feet, (b) 4 feet 
and (c) 10 feet from the centre? 



Varied Motion — Acceleration, 

22. Varied Motion. The motion of a body is said to 
be varied, and its velocity is called variable, if it moves 
through unequal spaces in equal successive intervals of 
time. The motion (supposed to be continuous) is said 
to be uniformly varied if the velocity (1) increases or 
(2) decreases by the same amount in equal successive 
intervals of time, however small these be taken. 



14 KINEMATICS. [23. 

In the first case the motion is uniformly accelerated, 
as the motion of a stone falling toward the earth; in the 
second case it is uniformly retarded, as that of a stone 
thrown vertically upward. * 

The velocity of a body, if variable, is measured at any 
instant by the distance through which the body would 
pass in the folloiving unit of time, if the motion -were to 
continue uniformly through that time at the same rate. 

Thus, we speak of the velocity of a railroad train as 
being at a certain instant 25 miles per hour, meaning 
that, if the rate were to be kept up uniformly for the 
hour following, the train would pass over 25 miles. We 
know, however, that the supposition will not in fact be 
realized. Again, the velocity of a falling body, at a 
certain instant, maybe said to-be 64 feet per second; 
and by this is meant that, if it should move uniformly 
for the next second at the rate it has at the instant 
under consideration, it would pass over 64 feet. But in 
fact its velocity is constantly increasing, and it will 
actually fall through a space greater than 64 feet. 

23. Acceleration. If the motion of a body is uni- 
formly accelerated, the equal increment of velocity for 
each succeeding unit of time — the second — is called the 
acceleration; it is the rate of change of velocity. 

For example, a body falling freely from rest toward 
the earth acquires a velocity of about 32 feet per second 
at the end of 1 second, at the end of 2 seconds its velo- 
city is (32) + 32, of 3 seconds it is (32 + 32) + 32, and 
so on. In other words, whatever the previous velocity 
it may have at any instant, in the second following its 
velocity is increased by about 32 feet. This increment 
of velocity of 32 feet-per-second per second (as it should 



25.] VARIED MOTION — ACCELERATION. 15 

be expressed in full) is called the acceleration due 
to gravity, and is denoted by the letter g. In general 
the acceleration due to the action of any force (as ex- 
plained in 60) is expressed by the letter/. 

It is explained in a following Art. (64, p. 61) that the value of 
g varies slightly for different points on the earth's surface, being 
greatest at the poles and decreasing toward the equator, where its 
value is least. The value for New York is about 32.16 (some- 
times called 32i). it also diminishes as the distance from the sur- 
face of the earth increases. 

It is also explained in article 65, p. 62, that this acceleration 
due to gravity is the same, at one place, for all bodies, whatever 
their mass; that is, a bullet and a feather will fall in the same time 
to the earth from a given point, and acquire the same velocity, if 
the resistance of the air is eliminated. (Read articles 64, 65, 250.) 

If the motion of a body is uniformly retarded, the 
term acceleration is also employed to indicate the equal 
loss of velocity for each succeeding second, but it has 
then a negative sign, as having a direction opposite to 
that of the initial velocity of the body. This is true of 
a body thrown up from the earth, or of a body projected 
on a rough horizontal plane and retarded by friction. 

24. Velocity acquired in Uniformly Accelerated Mo- 
tion. Since the acceleration (/) of a body is the incre- 
ment of velocity for each successive second, it is clear 
that, if the body starts from rest, its velocity (v) at the 
end of t seconds will be equal toft. That is: 

v = ft; for a falling body v = gt. 

25. Space passed over in Uniformly Accelerated Mo- 
tion. The space passed over by a body, starting from rest 
and moving with uniformly accelerated motion, is equal 



16 



KINEMATICS. 



[25. 



to one half the product of the acceleration into the square 
of the time. 

* = ift*; for a falling body s = igt 9 . 

Let AB (Fig. 5) be taken proportional to the time (t) 
and BO, at right angles to it, proportional to the velocity 
(v) acquired in this time, and connect AC; it will be 
shown that the space passed over by the body is repre- 
sented by the area of the triangle ABC. 




Fig. 5. 



First, it is necessary to show that if Ah represents any 
other time (t') in terms of the same unit, then the cor* 
responding perpendicular he represents the velocity (v ') 
acquired in this time. For (24) 

v = ft, and v' = ft f , 



Also, 



But, by supposition, Ah represents t', hence he must 





J t 


v' 




BC 
AB 


= % but 


BC 

AB "" 


V 

T 




he 

'"' Ah~ 


v 9 





•] 



YAEIED MOTION — ACCELEEATION. 



17 



represent v'\ that is, the corresponding velocity acquired 
in this time. 

Again, let the time (t) be divided into any number of 
equal parts represented geometrically by Ab', b'b n ', etc. 
(Fig. 6). Erect the perpendiculars b'c', b"c" , etc.; by 
the preceding paragraph, these perpendiculars will 
represent geometrically the velocities acquired at the 
end of these times taken from the beginning. Now 
suppose (1) that the body moves uniformly for each of 
these portions of time with the velocity it has at the 
beginning of that interval, and (2) with that acquired 




3" h"' i" & 

Fig. 6. 



at the end. That is, on the first supposition, it moves 
for the time Ab f with the velocity 0; for the time Vb" 
with the constant velocity b'q'\ for the time b"b"' with 
the velocity b"c", and so on. Then, by Art. 21, the 
sum of the interior rectangles O.Ab', b'e' (=#'#". b'c'), 
b"e", and so on, will represent the whole space passed 
over on this first supposition. 

On the second supposition the body moves for the 
time AV with the constant velocity Vc'\ for the time 
Vh" with the velocity l"c"\ for the time V'V" with 
the velocity 5"V", etc. Then, in this case, the total 



18 KINEMATICS. [26. 

space passed over will be the sum of the exterior rect- 
angles d'V (= AV X Vc% d"V, d"'V", and so on. 

It is obvious that the space represented by the sum of 
the interior rectangles is less than the true space passed 
over by the body, and that represented by the sum of the 
exterior rectangles is greater than the true space; and 
each differs, by a series of small step-like triangles, from 
the area of the triangle. Now if the number of intervals 
into which t is divided be increased indefinitely, and 
consequently the length of each be indefinitely dimin- 
ished, and the same construction as that above supposed 
be carried through, then the sums of the interior and 
exterior rectangles will approach the area of the triangle 
as their limit. But the spaces passed over by the body, 
upon the two suppositions made, also approach the 
true space (corresponding to a continual and unbroken 
increase in velocity) as their limit. But when two sets 
of variable quantities, which are always equal, simul- 
taneously approach their limits, these limits are equal. 
Therefore 

The true space is geometrically represented by the area 
of the triangle. 

.-. s = iBC.AB = \vt = \fi\ 

26. Average Velocity. From the preceding article 

s = \vt, and \v = -. This value of the velocity, ob- 
t 

tained by dividing the whole distance by the time, is 

called the average velocity. In the case of uniformly 

accelerated motion, the average velocity is equal to one 

half the final velocity acquired ; or, in other words, 

the space passed over is equal to one half the product of 

this final velocity into the time. This is represented 



27.] 



VAEIED MOTION — ACCELERATION. 



19 



geometrically by Fig. 7, where, if BC = 2BE, it is seen 
that the areas of the triangle ABC (= i.vt) and of the 
rectangle A BED (= \v.t) are equal. 

The term average velocity is also employed, in the case 
of varied motion in general, to denote the result ob- 
tained by dividing the whole space by the time. For 
example, if a train traverses 100 miles in 4 hours, its 
average velocity is said to be :L ^ Q - = 25 miles per hour, 

c 



t * 

Fig. 7. 



although its actual velocity may have varied through 
very wide limits during the time. 

27. Formulas for Accelerated Motion. The results of 
articles 24 and 25 give 

v = ft; for a falling body v = gt. (1) 

s = iff; " « « s = \gt\ (2) 

Therefore, eliminating t from (2), 




; for a falling body 




(3) 



v 2 = 2gs 



These three equations give the most important rela- 
tions for bodies starting from rest and moving with uni- 
formly accelerated motion. From them we see that 



20 KINEMATICS. [28. 

(1) The velocity acquired is proportional to the time. 

(2) The space is proportional to the square of the time. 

(3) The space is proportional to the square of the velo- 
city acquired. 

28. From equation (2) of the preceding article it is 
seen that the space described in the first second from rest 
is equal to one half the acceleration. Also, the spaces 
described in 1, 2, 3, 4, etc., seconds are, by the formula: 

lsec. 2. 3. 4. 

*/, if, if, W, etc. 

Therefore the spaces described in the first, second, third, 
etc., seconds will be: 

1st sec. 2d. 3d. 4th. 

if, if, if, if, etc. 

In other words, the spaces described in the successive 
seconds are proportional to the numbers 1, 3, 5, 7, 9, 
etc. ; for the n th second the space will be by this law 

— - — /, or for a falling body — - — g; this is equal to 

the space passed over in (n — 1) seconds [= if(n — l) 2 ] 
subtracted from the space passed over in n seconds 

EXAMPLES. 

II. Uniformly Accelerated Motion (Articles 22-28). A. Falling 
Bodies {take g = 32). 

[It is to be understood in each case that the body falls from 
rest, and that the resistance of the air is neglected. It is to be 
remembered, also, that the assumption that the value of g is con- 
stant for points above the surface of the earth is not rigidly true.] 

1. A body falls 15 seconds: Required (a) the velocity acquired; 



28.] VARIED MOTION — ACCELERATION. 21 

(b) the whole distance fallen through ; (c) the space passed over in 
the last second of its fall ; (d) the space in the last three seconds. 

2. A body has fallen through 5184 feet: Required (a) the time 
of falling; (b) the final velocity. 

3. A body has acquired in falling a velocity of 512 feet per 
second: Required (a) the time of falling; (b) the distance fallen 
through. 

4. A body in falling passed over 336 feet in the last second: 
Required (a) the time of falling; (b) the distance fallen. 

5. A body in falling passed over 1008 feet in the last three 
seconds: Required (a) the time of falling; (5) the distance fallen 
through. 

6. What is the ratio of the velocities of a falling body at the end 
of the first £, -£, 1, 3, and 4£ seconds ? Find the actual veloci- 
ties in this way, from the velocity at the end of 1 second (g). 

7. What is the ratio of the spaces passed over by a falling body 
in i, i, 1, 3, 4^ seconds ? Obtain the respective distances in this 
way, from that of 1 second (16 feet). 

8. A sand-bag is dropped from a balloon, which is for the 
moment at rest at a height of 3 miles: Required (a) the time of 
falling to the earth, and (b) the velocity acquired. 

9. What is the distance fallen through in the third of a second, 
commencing (a) the 6th second, and (b) the 11th second? 

10. Two balls are dropped at the same instant from points 100 
feet apart vertically: What distance will separate them at the end 
of 2, 3, and 5 seconds ? 

11. Two balls A and B are dropped from a height, B 2 seconds 
after the other : (a) How far apart will they be after B has fallen 
2, 3, and 5 seconds ? (b) When will they be 416 feet apart ? 

12. A stone is dropped down a well 224 feet deep : How soon 
will the splash in the water be heard at the top if the velocity of 
the sound is 1120 feet per second (corresponding to a tempera- 
ture of the air of about 60° F.)? 

13. A stone is dropped from the top of a cliff, and after 6| 
seconds it is heard to strike the ground below : How high is the 
cliff, taking the velocity of sound as 1152 feet per second (tei 
perature of air about 90° F.)? 



22 KINEMATICS. [29 



B. General Case. — Acceleration =/. 
[The motion is assumed to be uniformly accelerated.] 

1. A body moves 100 feet in the first 5 seconds from rest, 
What is the acceleration ? 

2. A body moves 10 feet in the first second : (a) What is the 
acceleration ? (b) How far will it go in 8 seconds ? (c) What win 
be its final velocity at the end of this time ? 

3. The acceleration is 12 feet-per-second per second: (a) What 
velocity does a body acquire in 6 seconds ? (b) What space does it 
pass over ? 

4. A body passes over 36 feet in the fifth second: What is the 
acceleration ? 

5. The acceleration due to the attraction of Jupiter for bodies 
on or near its surface is about 2.6 times g : (a) What velocity 
would a falling body acquire in 3 seconds ? (b) What space would 
it pass through in this time ? 

6. What time would be required in the above case (5) for a body 
to fall 2340 feet ? 

7. The acceleration of gravity en the moon is about \g : How 
long and how far must a body fall to acquire a velocity of 32 feet? 

8. The acceleration of gravity on the sun is about 28 X ff '• 
Compare the acquired velocities and spaces fallen through for 
the first three seconds with those true for the earth. 

9. A body moves 45 feet in 3 seconds, and 80 feet in the next 
2 seconds : Is its motion uniformly accelerated ? 

10. A body passes over 50 feet in 5 seconds: What distance 
must it go in the next 5 to satisfy the condition of uniformly 
accelerated motion ? 

Composition and Resolution of Velocities — Uniform 
Motion, 

29. Composition of Motions in General. It was ex- 
plained in Art. 11 that, when a body is said to be in 
motion, reference is always made to some other body 
with respect to which the first body changes its posi- 



31.] COMPOSITION OF MOTIONS. 23 

tion. In many cases which arise we have to consider 
not the simple motions of bodies, but their actual 
motions as composed of several different motions. For 
example, a man walking on the deck of a steamboat is 
in motion with reference to it, but the boat in turn is 
in motion as compared with the neighboring shore. 
Therefore his actual motion with reference to the land 
is composed of his own independent motion and that of 
the boat. Hence we have, in such cases, to do with the 
coexistence of motions; and the problem arises, when the 
separate motions are given, to find the actual resulting 
motion in rate (velocity) and direction. 

30. Resultant and Component Velocities. When a 
body tends to move at the same time with several 
different velocities, either in the same or different 
directions, the actual velocity due to the combination 
of all is called the resultant, and the separate velocities 
are called the components. The process of finding the 
resultant, when the components are given, is called the 
Composition of Velocities. 

31. Composition of Constant Velocities in the same 
Straight Line. The resultant of two component velocities 
in the same direction is equal to their sum; if they have 
opposite directions, it is equal to their difference. In 
general, if of several velocities those in one direction are 
called plus (+)> and those in the opposite are called 
minus (—), the resultant is equal to their algebraic sum. 

For example, a boat, moving uniformly at the rate of 
6 miles per hour down a stream running at the uniform 
rate of 4 miles, has a resultant velocity in the same 
direction of 10 miles (6 -f 4). If the boat is headed up 
stream and keeps the same rate, the resultant velocity is 



24 KINEMATICS. [32. 

also up stream and equal to 2 miles (6 — 4). If, in the 
latter case, the stream had a velocity of 8 miles, the 
resultant velocity would be equal to — 2, (6 — 8); that 
is, the boat would in fact drift down stream at this rate. 

32. Composition of two Constant Velocities not in the 
same Straight Line. If the two component velocities are 
not in the same straight line, then the resultant velocity 
will lie between them, and will be determined in direc- 
tion and magnitude by the Parallelogram of Velocities. 

33. Parallelogram of Velocities. This principle is 
stated as follows : If the component velocities be repre- 
sented in direction and mag7iitude by the two adjacent 
sides of a parallelogram, the resultant velocity will be 
given by the diagonal passing through their point of 
intersection. Suppose a body tends to move uniformly 







cpS 


t/j 




c '>^\ 




c£s 












V 


b" b'" M 

Fig. 8. 



from A toward B (Fig. 8) with a velocity u, represented 
by the line AB; also at the same instant from A to D 
with a velocity v, represented by AD, then the body will 
actually move in the direction A O with a constant velo- 
city represented by AC. (See Arts. 68, b, and 128.) 

For the body, if it had only the velocity u, would in 
one second move from A to B, and if only the velocity 
v, would move from A to D\ but the motion in the one 



34.] COMPOSITION OF MOTIONS. 25 

direction cannot effect that in the other if they go on 
together, so that at the end of the given time the bod}' 
will actually be at C, having moved along the straight 
line AC. But, again, if Ab', b'b", b"V", etc., be taken 
to represent the motion of the body in equal intervals of 
time in the direction AB with che velocity u, and Ad', 
d'd", d"d ,n , etc., the motion in the same intervals of 
time in the direction AD with the velocity v, the result- 
ant motion will be represented by Ac', c'c", c"c'", etc. 
But these distances are equal, since they are by similar 
triangles proportional to the equal distances Ab', b'b", 
etc. (or Ad', d'd", etc.); therefore the motion of the 
body in the resultant direction is also uniform. 

34. Calculation of the Magnitude and Direction of the 
Resultant Velocity. From trigonometry we have (see 
also Art. 130) 

AG 2 = AB' + AD' + 2AB.AD cos y. 
Therefore 

P = u u + v a + 2uv cos y. 

/ > ^ 




Also, in the triangle ABC (Fig. 9) the component 
velocities are represented by AB = u, and BC(= AD) = 
v; the resultant velocity is A C = V; also, BA C = a, 
ACB (= CAD) = p, and CBA = 180° - BAD = 180° 
— y = 180° — (a -f- /?). Hence the relations in direc- 



26 



KINEMATICS. 



[35. 



tion and magnitude of the resultant and component 
velocities may be calculated by the usual methods for 
the solution of this plane triangle, where three parts 
are given and the others required. 

It is further seen from this case that the relation of 
the two component velocities and their resultant may be 
expressed geometrically by the triangle ABC, hence 
sometimes called the Triangle of Velocities. 

35. For the special case (Fig. 10) where the directions 



2? 




t 


V 




y^ 



rf 



u 

Fig. 10. 



of the component velocities are at right angles to each 
other, the relations are more simple. Here 

V= tf v * _|_ u 2 ; also, cos a = sin j3 = — , 



sin a = cos/? 



v 
T' 



tan a = cot fi 



36. The following cases may be taken as illustrations of th<? 
Parallelogram of Velocities. — Suppose a boat to move uniformly 
at the rate of u miles per hour across a stream running at the rate 
of v miles per hour. Here the velocities u and v are the com- 
ponents, and, if Ah, Ad be taken to represent them in their 
respective directions (Fig. 11), the resultant velocity will be given 



37.] 



COMPOSITION OF MOTIONS. 



27 



in direction and magnitude by the diagonal Ac. Hence the boat 
will actually move in the direction Ac, (tan a = — 1 at the rate of 

V miles an hour (= VV-j- -w 2 ), and will reach G in the same time 
in which it would have gone without the current directly across 
to B, or would have drifted with the current to D. 

Again, suppose the component velocities as above; if it be 
required that the boat shall go directly across to G, here the direc- 
tion of the resultant is AG (Fig. 12), and the components are 



SI 



/.••• 



& 



B 



Fig. 11. 



Fig. 12. 



Ab (= u) and Ad (= v). Hence the boat must be headed up 
stream at an angle BAGAsin a = — I, and the resultant velocity 
will be expressed by ^v? — xiK 

37. Composition of several Constant Velocities. If 

there are more than two component velocities, then the 
method of finding their resultant is as follows: Find the 
resultant of two of the component velocities, then of this 
resultant and the third component, again of the last 
resultant and the fourth component, and so on. Thus 
let AB, AC, AD, AB (Fig. 13), represent several com- 
ponent velocities; the resultant of AB and AC is (33) 
the diagonal Ac; again, the resultant of Ac and AD, that 



28 KINEMATICS. [38. 

is of AB, AG, AD, is Ad; still again, the resultant of 
Ad and AE, that is of the four original velocities, is Ae. 
It will be seen by comparing Figs. 13 and 14, that the 
sides of the polygon ab cde (Fig. 14) represent the 
four velocities and their resultant. Hence, in general, 
if the component velocities be laid off in order of direc- 
tion, as ab, be, cd, de (Fig. 14), the side which com- 



Fig. 13. Fig. 14. 

pletes the polygon so formed, viz. ae, represents the 
resultant velocity. This is sometimes called the Poly- 
gon of Velocities, 

38. Resolution of Velocities. The process of finding 
the component velocities, which shall be equivalent to a 
given resultant velocity, is called the Resolution of Velo- 
cities. If the required components are two in number 
and their directions are given, then their magnitude is 
found by completing the parallelogram whose diagonal 
is the given resultant velocity and whose sides have the 
given directions. 

Let (Fig 15) A C represent the resultant velocity, and 
let AX and A Y be the directions of the required com- 
ponents; draw through C the lines CD, OB parallel 
respectively to AX and AY, then AB and AD, the 



38.] 



COMPOSITION" OF MOTIONS. 



29 



sides of the parallelogram thus formed, will represent 
the required components in direction and magnitude. 





Fig. 15. 



Fig. 16. 



If the directions AX and AY are at right angles 
(Fig. 16), and a represents the angle CAB, then the 
components are AB = AG cos a and AD = A G sin a. 

For example, suppose a boat to move uniformly in the 
direction AG (Fig. 17) in virtue of its own motion 



A 



SI 



Fig. 17. 



directly across the stream and the velocity of the cur- 
rent toward D, taken together, both being uniform. 
Then if Ac represents this resultant velocity, the com- 
ponent velocities will be given by Ab (= Ac cos a) and 
Ad (= Ac sin a). If the transfer across the stream were 
alone desired, the component Ab in this direction might 
be called the effective velocity. 



30 KINEMATICS. [38. 



EXAMPLES. 

III. Composition of Velocities. Articles 29-37. 

[The velocities are supposed to be constant in all cases.] 

1. The velocity of a steamboat is 5 miles per hour, that of the 
stream is 4 miles, and a man walks the deck from stern to bow at 
the rate of 3 miles : Required the actual velocity of the boat (a) if 
headed up stream, and (b) down stream; also (c, d), that of the man 
in each case. 

2. The velocities of boat and stream are as in example 1, and 
the boat is headed directly across the stream (Fig. 11): (a) What 
will be the actual direction of the boat's motion ? (b) What the 
rate of its motion? (c) How long will the passage take if the 
stream is 2 miles wide ? (d) Where will the boat land (BG = ?). 

3. The velocities of boat and stream are as in 1 and 2, but it is 
required that the boat shall go directly across from A to C (Fig. 
12): (a) In what direction must the boat be headed? (b) What 
will be its actual velocity across ? (c) What will be the time of 
passage, the width being 2 miles ? 

4. Find answers for the three questions in example 3, on the 
supposition that the boat is to reach a point 30° up stream from 
the starting-point. 

5. Find answers for the three questions in example 3, on the 
supposition that the boat must reach a point 30° down stream. 

6. The velocity of the boat is 4 miles per hour, and that of the 
stream 5 miles ; the width of the stream is 1 mile : What is nearest 
point, to that directly across from the starting-point, which the 
boat can reach ? (Solve this problem by geometrical construction.) 

7. A ball on a horizontal surface tends to move north with a 
velocity of 12 feet per second, and east with a velocity of 5 feet 
per second : (a) What will be the actual velocity, and (b) in what 
direction ? 

8. A ball, moving north at a rate of 8 feet per second, receives 
an impulse tending to make it move due north-east with the same 
velocity: (a) What path will it take, and (b) at what rate will it 
move ? 

9. A man, skating uniformly at a rate of 12 feet per second, 



38.] COMPOSITION OF MOTIONS. 31 

projects a ball on the ice in a direction at right angles to his 
motion at a rate of 9 feet per second : What is (a) the actual rate, 
and (b) the direction of its motion (friction neglected)? 

10. A ball tends to move north at the rate of 8 feet per second, 
also S. 60° E. and S. 60° W., each at the same rate : What is its 
actual velocity ? 

11. A ball tends to move east 5 miles per hour, also N. 45° W. 
and S. 45° W., each at the same rate: Required the direction and 
rate of motion. 

12. If a boat headed directly across a stream moves at the uni- 
form rate of 100 yards a minute, while the current runs 80 yards 
a minute : (a) In what direction will it actually go, and (b) what 
distance will it land down stream ? (c) What should be its course 
in order that it may reach a landing-place directly opposite the 
starting-point, and (d) how long would the passage take ? The 
width of the stream is 1200 yards. 

IY. Resolution of Constant Velocities. Article 38. 

1. A ball tends to move in a certain direction at a rate of 9 feet 
per second, but it is constrained to move at an angle of 30° with 
this direction : Required its velocity in the latter direction. (Fig. 16.) 

2. A body moves uniformly about a semi-circumference at the 
rate of 12 feet per second: What is the component of its velocity 
parallel to the diameter when it is 30°, 60°, 90°, 120°, and 180° 
from the starting-point? (Fig. 16) 

3. A ball rolls at the rate of 8 feet per second across the diagonal 
of a rectangular room ABGD whose dimensions are 15 X 20 
(= AB X AG): What is its rate of motion parallel to each side? 

4. A body moves N. 30° E. at a rate of 6 miles per hour: 
Required its rate of motion northerly and easterly? 

5. A boat, though headed directly across a stream, actually 
moves diagonally across the stream at an angle of 30° {B A G, Fig, 
11, p. 27), and at a rate of 10 miles per hour: Required (a) the rate 
of the boat, and (b) of the current, each taken independently. 

6. A boat steams directly across a stream 1800 yards wide in 
30 minutes, the current flowing all the time at the rate of 80 yards 
per minute : What would be the direction and rate of motion if 
there were no current ? 



32 



KINEMATICS. 



[39. 



7. A balloon has a velocity of 20 feet per second in an upward 
direction which makes an angle a with a vertical line : If its ve- 
locity vertically upward would be 1000 feet per minute, what is 
its horizontal velocity due to the wind ? What is a ? 



Composition and Resolution of Accelerations. 

39. Composition and Resolution of Accelerations. The 

composition and resolution of velocities may be extended 
also to the case of uniform accelerations, the method 
being in all respects similar to that in the preceding 
articles. The sides of the parallelogram here represent 
the component accelerations, and the diagonal the re- 
sultant acceleration. 

The simplest application of the principle of the reso- 
lution of accelerations is to the case of motion down an 
inclined plane (40). 

40. Motion down an Inclined Plane. The direction of 
the acceleration of gravity is that of a vertical line, and 




Fig. 18. 

a body falls in this direction if entirely free; but a body 
on an inclined plane is only free to slide along it, and the 
acceleration is here that component of the whole accele- 
ration which is parallel to the plane; viz., g sin a. 

Let (Fig. 18) ac be taken to represent the vertical 
acceleration g; the directions of its components are 
respectively parallel and perpendicular to the plane, and 



40.] COMPOSITION OF MOTIONS. 33 

are represented by ab and ad. But lac = HLK — a, 
and therefore ad = fo = «c sin <#. That is, a<#, or the 
acceleration down the plane, is equal to g sin a. 

The formulas of Art. 27, for a falling body, are then 
applicable to the case of a body sliding down a smooth 
inclined plane, if for g we write g sin a. That is : 



V 


= 


# sin « . 


t, 


s 


= 


%g sin a 


•t% 


V* 




2# sin a 


. s. 



In the last formula, if the body descends from if to L, 
s = i/X, and s sin <* = iTX or h, the height of the 
plane; 

From this equation it follows that: the velocity acquired 
in descending any inclined plane is the same as that 
gained in falling through the vertical height of the plane. 
This is also true for a continuous curve. This principle 
finds an application in Art. 244. 

EXAMPLES. 

V. Falling down an Inclined Plane. Article 40. 

[The plane is supposed to be perfectly smooth, so that there is no 
friction.] 

1. The angle of the plane is 30°: Required (a) the acceleration 
down the plane; (b) the distance fallen through in 4 seconds; (c) 
the velocity acquired; (d) the distance in the last second. 

2. The height of the plane is 100 feet and the length 400: (a) 
What is the time required to reach the bottom ? (b) "What is the 
velocity acquired ? 

3. The angle of the plane is 45° : Required the time of falling 
144 feet. 



34 KINEMATICS. [41. 

4. The length of a plane is 576 feet, a body falls down it in 24 
seconds: (a) What is the acceleration ? (b) What is the height of 
the plane ? 

5. The height of a plane is 98 feet, and a body gains a velocity 
of 20 feet per second in falling 5 seconds on it: Required (a) the 
acceleration ; (b) the length of the plane. 

6. The height of a plane is 256 feet, a body reaches the 
bottom in 16 seconds: (a) What is the length of the plane ? (b) 
What is the velocity acquired ? 

7. Several planes, having the same altitude, viz. 400 feet, have 
lengths 600, 800, 1200, and 1600 feet: Compare the times of 
descent and acquired velocities for each. 

8. Show that for several planes having the same altitude the 
times of descent are proportional to the lengths ; that is, 

* l 4 7 

-p = -j or t oc I. 

9. Prove that the time of falling from rest down a chord of a 
vertical circle, drawn from the highest point, is constant. 



Composition of Uniform and Accelerated Motion in the 
same Line. 

41. Composition of Uniform and Accelerated Motion in 
the same Straight Line. (Only the cases of uniformly 
accelerated and retarded motion will be considered.) 

If of two component velocities one is constant (u) 
and the other is uniformly increasing — that is, tending 
to produce uniformly accelerated motion — but both in 
the same line, then the resultant is equal to their sum 
or difference according as they have the same (a) or 
opposite directions (b). 

(a) In the first case, represent the uniform velocity 
by u, and that produced by the accelerated motion by 
v (= ft); then, if Fis the resultant velocity, 

V = u + v = u -\-ft; for a falling body V = u -f gt. (1) 



42.] COMPOSITION OF MOTIONS. 35 

The? last formula applies to the case of a body pro- 
jected \rith an initial velocity vertically downward toward 
the surface of the earth from a point above. The 
resultant velocity is the sum of this initial velocity and 
that due to its accelerated motion caused by gravity (24). 

(#) In the second case 

V = u — v = u — ft; f or a falling body V = u — gt. (2) 

The last formula here applies to the case of a body 
projected vertically upward from the earth with an 
initial velocity u; its resultant velocity at any moment 
is then equal to the initial velocity diminished by the 
velocity due to the accelerated motion downward (that 
is, in the opposite direction) caused by gravity. 

42. The distance (s) which a body passes over in a 
given time, in the above examples, is to be found by tak- 
ing the sum, in the first case, and the difference, in the 
second, of the space that would be passed over if it 
moved uniformly for the time t (19) with the velocity u, 
and that it would pass over independently in the same 
time in consequence of the accelerated motion (27). 

Therefore (a) 

s = ut + iff; for a falling body s = ut + igf. (3) 
And (b) 

s = ut — iff; for a falling body s =ut — igf. (4) 
By combining equations (1) and (3), since V — u -{-ft, 
V* = u 2 + 2uft + f*f=u* + 2f{ut + i ff) = tf+Zfs* 
Therefore 

V 2 = if + 2fs; for a falling body V*=u* + 2gs. (5) 
In the same manner 
F 3 = u* - 2fs ; for a falling body V 2 = u* - 2gs. (6) 



36 



KINEMATICS. 



[43. 



43. Geometrical Representation. It was shown, in 
Art. 21, that the space passed over by a body moving 
uniformly may be represented geometrically by a rect- 
angle; and again, in Art. 25, that the space described 
by a body moving with uniformly accelerated motion 
may be represented by a right-angled triangle. If now 
a body has an initial velocity in the same or opposite 
direction to that in which it begins to move with uni- 
formly accelerated motion, the space passed over will be 
represented by a geometrical figure formed by the com- 
bination of the rectangle and triangle. 

For example, in Fig. 19, let AB represent the time 
(t), BC the initial velocity (u), also CE the velocity (v) 
acquired in this time and in the same direction as u; 
then will the whole space passed over be represented by 
the figure ABED, which is the sum of the rectangle 
ABCD (ut) and the triangle DOE ($vt = iff). 





Again, suppose the accelerated motion to be in a 
direction opposite to that of the initial velocity. Let 
AB (Fig. 20) represent the time (t), and BC the initial 
velocity (u), also take CE to represent the velocity 
acquired (v) in the given time; then the space described 
will be proportional to the area of the quadrilateral 



44.] COMPOSITION OF MOTIONS. 37 

ABED, which is the difference between the rectangle 
ABCD (ut) and the triangle DCE (ivt = iff). 

44. Motion of a Body projected vertically upward. 

The three formulas obtained in Art. 42, which give the 
relations of the velocity, space, and time of a body which 
has an initial Telocity in a direction opposite to that in 
which it tends to move with accelerated motion, have 
an especial importance. They are : 

V=u-ft, s = ut-ift*, V 2 = u*-2fs. 

For a falling body these are : 

V=u- gt, (1) s=ut- \gt\ (2) F 2 = w a - 2gs. (3) 

The relations given below are deduced from the last 
three equations, since the case of the body projected 
vertically upward is practically the most important, but 
all the results obtained may be made general by writing 
/ for g. 

1. The Time of Ascent. From equation (1), if t — 

— that is, at the moment of starting — V = u, the initial 

velocity; as t increases V diminishes, and when gt = u, 

u 
or t = -, then V = 0. That is, at a time after the 

g 

u 
starting, expressed by t = — , the body will for an instant 

come to rest. 

2. Time of Descent. If in the same equation gt is 

greater than u, i.e. t is greater than — , the value of V 

will be negative; in other words, the body will begin to 
descend. When it reaches the starting-point again, 
s = 0, and therefore, from equation (2), 

tit — \gf = 0, and t = or — . 



38 KINEMATICS. [44. 

The value t = corresponds obviously to the moment 

of starting, and t = — means that at the end of this 

time the body will have returned to the starting-point. 

The time of ascent and descent is then — ; and since 

9 
u 
the former = — , the time of descending must be also 

equal to — . 
9 

3. Height of Ascent. At the highest point reached 
V = 0, and therefore in equation (3) 

= w 2 — 2gs, and s = - 7 — , 

and this value gives the distance ascended. 

This can also be obtained from equation (2); for at 

the highest point t = — , therefore 
_ w a w 2 __ v? 

This equation is the same as (3) in Art. 27; hence the 
result here obtained may be stated in this form: A body 
projected vertically up will ascend to a height from 
which it must fall to acquire a velocity equal to that of 
its projection. 

4. Velocity acquired in descending. The time re- 
quired for the whole ascent and descent is — , therefore 
in equation (1) 

V = u — a . — = u — 2u = — u y 
y 9 



45.] COMPOSITION OF MOTIONS. 39 

or the velocity acquired in descending is equal to the 
initial velocity, but in the opposite direction. 
If in equation (3) we let s = 0, then 

V* = U\ V= +UOT -U, 

which result corresponds to that just given. 

Furthermore, for any value of s there will be two 
different values of t from equation (2), corresponding to 
the time when it passes the given point on the ascent, and 
that when it returns to it on the descent. Also, at any 
point on the descent the velocity (equation 3) will be 
the same with the contrary sign as that on the cor- 
responding point in the ascent. 

45. Projected up or down an Inclined Plane. For a 

body moving up or down a smooth inclined plane with 
an initial velocity u, the relations are the same as those 
given in articles 41, 42, and 44, except that, as in 40, in 
every case we must write g sin a for g. 

Down. Up. 

V = u -f- g sin a. t, V = u — g sin a.t, 

s = ut -f- ig sin a.f, s = ut — \g sin a.f, 

V* = y? + 2g sin a.s. V 2 = v? — %g sin a.s, 

EXAMPLES. 

VI. Bodies projected vertically downward. Articles 41, 42. 
[The resistance of the air is neglected.] 

1. A body is thrown vertically down with an initial velocity of 
36 feet per second : Required (a) the velocity at the end of 7 sec- 
onds ; (b) the distance fallen through ; (c) the space passed over in 
the last second. 

2. A body is projected down with an initial velocity of 20 feet 
per second : (a) How long will it require to fall 594 feet ? (b) "What 
velocity will it then have ? 



40 KINEMATICS. [45, 

3. What velocity of projection must a stone have to reach the 
bottom of a cliff 270 feet high in 3 seconds ? 

4. With what velocity must a stone be thrown down the shaft 
of a mine 556 feet deep in order that the sound of its fall may be 
heard at the top after 4| seconds ? The velocity of sound is to be 
taken as 1112 feet per second. 

5. A body projected vertically down has a velocity of 215 feet 
per second at the end of 5 seconds: Required (a) the velocity of 
projection ; (b) the distance gone through. 

6. A body projected down passes over 133 feet in the fourth 
second: Required the velocity of projection. 

7. A stone is dropped from a bucket which is descending a 
shaft at the uniform rate of 12 feet per second, and at the moment 
when the bucket is 238 feet from the bottom : (a) How far will 
they be apart in 2 seconds ? (b) When will the stone reach the 
bottom ? 

8. A sand-bag is dropped from a balloon which is descending at 
the uniform rate of 24 feet per second ; after 8 seconds it strikes 
the ground : (a) What was the height of the balloon ? (b) How far 
were they apart after 5 seconds ? 

YII. Bodies projected vertically upward. Article 44. 

1. The velocity of the projection upward is 288 feet : Required 
(a) the time of ascent; (b) of descent; (c) the height of ascent; (d) 
the distance gone in the first and last seconds of ascent. 

2. A body is projected up with a velocity of 192 feet per 
second : (a) When will it be 432 feet above the starting-point ? (b) 
When will it be 720 feet below the starting-point ? Explain the 
double answer in each case. 

3. A body is projected up with a velocity of 208 feet: How long 
after starting will its velocity be (a) -\- 64, also (b) — 64 and (c) 
— 272 ? (The minus sign indicates downward motion.) 

4. What velocity of projection must a ball have in order to 
ascend just 900 feet ? 

5. What time does a body require to ascend 2304 feet, that 
being the highest point reached ? 

6. What velocity of projection is needed to make a body ascend 
just 6 seconds ? 



45.] COMPOSITION OF MOTIONS. 41 

7. A ball thrown up passes a staging 96 feet from, the ground at 
the end of 1 second : (a) What was the velocity of projection ? If 
the time is 6 seconds (b), what is the answer ? 

8. A body projected up passes over 112 feet in the fifth second of 
its ascent: "What was its velocity of projection ? 

9. A and B are two points 40 feet apart in a vertical line; a ball 
is dropped from A, and at the same instant one thrown up from 
B with a velocity = 80 feet per second : When and where will 
they pass each other ? 

10. A ball is dropped from the top of a cliff, and at the same 
instant another is thrown up with a velocity of 176 feet per 
second: (a) If they pass each other at the end of 2£ seconds, how 
high is the cliff ? (b) How far were they apart at the end of 2 
seconds ? (c) at the end of 3 seconds ? 

11. A ball is thrown up from the ground with a velocity of 128 
feet per second, and 2 seconds later another is thrown with a 
velocity of 160 feet: When and where will they pass each other ? 

12. A bucket is ascending a shaft uniformly at a rate of 32 feet 
per second: What will be the apparent motion of a stone dropped 
from it (a) to a person in the bucket ? (b) to a person on the side of 
the shaft opposite the initial point ? 

13. A balloon is rising uniformly at the rate of 96 feet per 
second ; at the instant it is 640 feet from the ground a sand-bag is 
dropped from the car: What will be the motion of the bag, 
when will it reach the ground, and over what distance will it 
have passed ? 

VIII. Projected up or down a smooth Inclined Plane. Article 45. 

1. The height of the plane is 114 feet, the length is 456, the 
velocity of projection down is 25 feet per second : (a) How long 
will it require to descend ? (b) What will be the final velocity ? 

2. The height and length are 144 and 576 feet respectively: (a) 
What velocity of projection up is required that it may just reach 
the top ? (b) What time will it take ? 

3. The angle of the plane is 30°, the velocity of projection down 
is 45 feet: Required (a) the velocity at the end of 4 seconds; (b) 
the distance gone through ; (c) the distance in the last second. 

4. The angle of the plane is 30°, the velocity of projection up is 



42 KINEMATICS. [46. 

80 feet per second: Required (a) the length of time the body will 
continue to go up, (6) the distance gone, and (c) the velocity at the 
end of 2 and of 8 seconds. 

IX. Bodies projected against Friction. Articles 41, 42. 

[The retardation (or minus acceleration) due to friction takes the 
place of the/ in the formulas of articles 42 and 44.] 

1. A body projected on a rough horizontal plane has at starting 
a velocity of 120 feet per second, but loses this at the rate of 19 
feet for each succeeding second: (a) What is the retardation 
(minus acceleration) due to friction ? (b) When will the body 
stop ? (c) How far will it have gone ? 

2. The retardation due to friction is for each second 8 feet pel 
second for a given sliding body, the initial velocity is 40 feet pet 
second: Required (a) the time it will continue to slide; (b) the 
distance it will go ; (c) its velocity at the end of 3 seconds. 

3. A railroad-car, when the engine is detached, has a velocity 
of 15 miles per hour, the retardation due to friction is 1 foot-per- 
second per second : How far and how long will the car continue 
to move ? 

4. If the retardation of friction is 4 feet-per-second per second: 
(a) What initial velocity (in miles per hour) must a body have in 
order to slide just 968 feet ? (b) If the velocity is doubled, how 
much farther will it go ? 

5. A body is projected up a rough inclined plane at an inclina- 
tion of 30°, the retardation of friction alone is 4 feet-per-second 
per second: If the initial velocity is 400 feet per second, how far 
and for how long will the body ascend ? 

Composition of Uniform and Accelerated Motion not in 
the same Straight Line. 

46. Composition of Uniform and Accelerated Motion. 

If a body tend to move in one direction with uniform 
motion, and in another direction with accelerated mo- 
tion, its actual path is not a straight line, but a curve. 
The same principle involved in the Parallelogram of 



47.1 



PEOJECTILES. 



43 



Velocities (33) makes it possible to determine the posi- 
tion of the body at the end of any given time (read Art. 
68, #, p. 68). For if one side of the parallelogram repre- 
sents, in direction and amount, the uniform motion in 
the given time, and the adjacent side the correspond- 
ing accelerated motion, the diagonally opposite point of 
the parallelogram will indicate the actual position of the 
body at the end of this time. In this statement nothing 
is said about the path which the body has described. 

47. Projectile. The simplest application of the above 
principle is to the case of the projectile. It will be 

>/ 




FK5.21. 

shown that, if the resistance of the air be neglected, the 
path of a projectile is a parabola. 

Suppose a body starts from A in the direction AD 
(Fig. 21) with an initial velocity u\ at the end of t 
seconds, if no other motion were imparted to it, it would 
reach a point D, so that 

AD = ut. (1) 



44 



KINEMATICS. 



[48. 



But from the instant of starting it falls vertically 
downward under the influence of gravity, with uni- 
formly accelerated motion. At the end of t seconds, if 
it had no initial velocity, it would fall to B, so that 

AB = igt\ (2) 

But as shown above (46), as the body must obey both 
tendencies to motion simultaneously, its actual position 
at the given time will be afc C. 

Squaring (1) and dividing by (2), we have 

AIT_BC^ uW _ 2u? 

AB ' ■ AB "" igt* ~ g ' 
or 

2w 3 

9 ' 



AB 

BO 9 
AB 



Therefore the ratio of the square of the ordinate BO 
(B'C, B"C") to the abscissa AB (AB', AB") is con- 
stant, and hence the curve is a parabola. 

48. Position of the Directrix, Axis, Focus. The line 
AD (Fig. 22) is a tangent to the parabola at A, the verti- 




cal line BAB is a diameter 

■. , BC* f 2u 
ratio of —:-=: 
AB \ 

to the directrix or to the focus 



9 I 



Fig. 22. 

The constant value of the 

is four times the distance from A 

The same relation is at 



49.] 



PEOJECTILES. 



45 



once obvious in the case of the parabola, whose equation 
is y 2 = 4:ax ( i.e. — - = 4a) ; it may also be proved analyti- 
cally for this case, where the co-ordinates EB, AD are 
oblique. It is proved geometrically in a following para- 
graph (50, e). 

u 2 
If AE is taken on the vertical line equal to ^-, and 

EGE' be drawn horizontally, this line will be the direc- 
trix; and if from A on the line A F (drawn so that the 

angle EAD = DAF) we take AF = ^-, the point i^is 

the focus of the parabola. The vertical line GML is 

the axis. 

If the direction of the initial velocity be horizontal, as 

in Fig. 23, then the starting-point 

A is the vertex, the vertical line AB 

is the axis, .and the focus and direc- 

u* 
trix are at distances equal to — from 

A, This figure shows well, as does 
also Fig. 21, what is meant by the 
statement in (27), that in uniformly 
accelerated motion the space is pro- 
portional to the square of the time. 
cessive intervals of time are equal, 




• *c* 

Fig. 23. 

Here, if the suc- 



and 



AD : AD' : AD" : AD"', etc., = 1:2:3:4, 



1 : 4 : 9 : 16. 



AB : AB' : AB" : AB"', etc., 

49. The actual path of a projectile deviates widely 
from a parabola because of the resistance of the air, 
which is very great with high velocities, as that of a 
cannon-ball or rifle-bullet (perhaps 1600 feet per second 



46 



KINEMATICS. 



[50. 



in starting). For this reason the maximum distance is 
gained, not by an angle of 45° (as shown below), but for 
an angle of a little over 30°. 

I A jet of water illustrates the subject of the projectile 
well, since each particle may be considered as an inde- 
pendent projectile, and thus the shape of the jet gives 
the continuous path. It shows, moreover, the deviation 
caused by the resistance of the air. 

50. Time of Flight, Range, etc. In Fig. 24 the angle 




•ff £ JT 3' X 

Fig. 24. 

HAK is called the angle of projection, and the horizon- 
tal distance AK is the range. 

(a) From the triangle A HE, AH ' = ut and HK = 
igf; also, 

SK _ igf_ gt_ 
2u' 



sin a 



\ t 



AH ut 

2u sin a 



a) 



The value of t in (1) gives the time of flight. 
(b) Again, 

AK = AH cos a — ut.cos a; 

or, substituting the above value of t, 



50.] 



PKOJECTILES. 



47 



AK 



2u* sin a cos a u 2 sin 2 a 



m 



This value AK gives the range. Further, since sin 2a 
= sin (180° — 2a) = sin 2(90° - a), it is obvious that 
for every horizontal distance there are two values of the 
angle of projection; viz., a and (90° — a). This is indi- 
cated in Fig. 25. The time of flight for the angle a is 

2u sin a .. , /rt/xo v .. . 2w cos or 

, and for (90 — a) it is . 

9 K ' 9 

The maximum range is obtained when a = 45° and 




sin 2a = 1, for the value of AK is then the greatest; 
for this case the two paths of the projectile coincide. 

(c) Since the vertical component of the initial velocity 
is u sin a, the actual vertical velocity of the projectile 
will be given, for any time t, by the formula (41) 

V = u sin a — gt. (3) 

For the highest point V = 0, and hence t = , and 

combining this with (1), it is seen that the times of 
ascent and descent are the same. 

(d) The distance, GB (Fig. 24), of the projectile above 



48 



KINEMATICS. 



[50 



the horizontal line AK is given, for any time t, as fol- 
lows: 

CB = DB - DC = ut.sina - \gt\ (4) 

For the highest point (M) t == , and hence 



mjst 



u sin <z 



(5) 



This value is greatest when sin a = 1 and « ±= 90°, in 
which case the parabola becomes a donble straight line. 
All the above results might have been obtained [as 
was (3) indeed] by the formulas in articles 41, 42, only 
taking u sin a for u, s being the distance above or below 
(— s) the horizontal plane. 




Fig. 26. 



(e) We may prove geometrically the point mentioned 
in Art. 48 ; namely, that the distance from A to the 

-w 2 
focus is equal to — ; that is, to one fourth of the con- 
stant value of the ratio of the square of the ordinate to 
the abscissa 



(B<J 



"~ 97 



\AB g 

Draw the line AF (Fig. 26) so that the angle DAF= 
BAE\ then, by the properties of the parabola, the focus 



51.] 



PEOJECTILES. 



49 



must lie in this line; it must also lie on the axis GL, 
and hence will be at F f their point of intersection. Now 



AL 



AL 



AL 





cos FAL 


" cos (90°- 


-2a) 


sin 2a 


From 


equation (2) above, 








AL = 


IAK- U * 


sin 2 a 
2g > 






,\AF = 


u 2 sin 2a . 


■ sin 2 a, 




or 


AF = 







The line EOF' is the common directrix of all the 
parabolas described by projectiles having the same ini- 
tial velocity but different angles of projection. The foci 
of all these parabolas lie on the circumference of a circle 



having A as its centre and a radius equal to 






51. The theory of the projectile may be further illustrated by 
the case of a jet of water flowing from a lateral orifice in the ver- 
tical side of a reservoir (Fig. 27). By 
a principle of hydrostatics the initial %__ . 
velocity of flow is the same as that 
which would be gained in falling 
freely through the height from the 
top of water to the orifice (27) ; that is, 

*= Y2^AG, or AC=^-. 

Supposing now that the level of 
the water is kept uniform, the direc- 
trix will coincide with it, that is, 
AK ; the focus will be at F, so that 

CA = CF = -g-. Further, it may be readily shown that the 




50 KINEMATICS. [51. 

range DE is the same for any two points taken, as Cand 0', so that 
AG = CD, and finally that the maximum distance DH is gained 
by an aperture in the middle at B, and is equal to AD(=2AJB). 



EXAMPLES. 
X. Projectiles. Articles 47-51. 

[The resistance of the air is left out of account.] 

1. The initial velocity of a projectile is 160 feet per second, and 
the angle of elevation is 30°: Required (a) the time of flight; (5) 
the range; (c) the highest point reached. 

2. When will the ball in example 1 be 96 feet above the ground? 
Explain the double answer. 

3. The initial velocity is 320 feet per second: What angle of 
elevation will give a range of 800 feet? Show that there are two 
answers. 

4. The angle of elevation is 15°: What initial velocity is re- 
quired that the range should be .4 miles ? 

5. A rifle-ball is shot horizontally from the top of a tower 100 
feet high, and with an initial velocity of 1200 feet per second: 
When and how far from the base of the tower will it strike tho 
horizontal plane below ? 

6. A ball is thrown horizontally from the top of a cliff above 
the sea; it strikes the water in 5 seconds and at a horizontal dis- 
tance of a mile: What was (a) the initial velocity, and {b) what 
was the height of the cliff ? 

7. If (Fig. 27) apertures are made at two points 36 feet from 
the top and bottom of the reservoir respectively, the whole height 
being 136 feet, what will be (a) the horizontal distance reached by 
the water in each case, and what {b) the initial velocity ? 

8. A stone is dropped from the top of a railroad-car, 16 feet 
above the ground, and when it is moving at the rate of 45 miles 
per hour: What will be its apparent motion (a) to a person on the 
train, (b) to one standing by the track ? (c, d) When and where 
will it reach the ground ? 

9. At what angle of elevation must a projectile be fired in order 
that it may strike an object 2500 feet distant on the same hori- 
zontal plane, the velocity of projection being 400 feet per second ? 



mm 



CHAPTER II.— DYNAMICS. 

52. The preceding chapter was devoted to the sub- 
ject of Kinematics, or the discussion of the motion of 
bodies without reference to their mass or to the force or 
forces which cause the motion. These latter subjects, 
included under Dynamics, or Kinetics, are considered 
in the present chapter. 

Mass — Density — Volume — Momentum. 

53. Mass or Quantity of Matter. The mass of a tody 
is the quantity of matter it contains. 

The relation in mass or quantity of matter, of differ- 
ent bodies of the same substance, and of uniform 
density (as defined in 56), is obviously given by the 
ratio of their volumes. For example, the mass or 
quantity of matter in a hundred cubic feet of iron is 
ten times that in ten cubic feet. For bodies of uniform 
density then: the mass is proportional to the volume. 

In general, however, for bodies of different substances 
it is possible to compare their masses only as the effect 
of a known force upon them is observed. Thus, we 
judge roughly as to whether a barrel is empty or full, 
and, in the latter case, as to the nature of the contents 
by noting the degree of resistance which it offers to a 
force tending to move it; e.g., a push or a kick. Simi- 
larly, if a ball of wood and another of the same size, of 
lead, attached to strings of equal length, be whirled 



52 DYNAMICS. [54. 

around at the same rate, the pull of the lead upon the 
centre will be the greater, and we form a rough estimate 
as to the relation of mass in this way. Could the pull at 
the centre be exactly measured under precisely the same 
conditions in each case, by means of a spring, the result 
would give the true relation of mass. 

Still, again, could the velocities given by the same 
force to two bodies in equal times be exactly determined, 
their ratio would give also the ratio of the masses of the 
bodies. No one of these methods of estimating the 
mass can be conveniently employed in practice. 

54. Mass determined by Weight. The simplest and 
at the same time most accurate method of comparing 
the masses of two bodies is by their weight, for the 
weight, or measure of the earth's attraction upon them 
determined by the balance, is, as proved by various ex- 
periments, proportional to the mass. Take two bodies 
of the same material, as two lumps of lead: if the weight 
of the first is twice that of the other, then it is easy to 
see that its mass, or the quantity of matter it contains, 
is also twice as great. But this is true in general: of 
two lots of lead and cotton, the bulk or volume of the 
latter may be much greater than that of the other, but 
if they have the same weight they have also the same 
mass; and if the weight of the lead is ten times that of 
the cotton, its mass is also ten times greater. 

Mass may be measured then by weight, and, in ordi- 
nary language, the latter word, expressed, for example, 
in pounds, is used as standing for the mass. In this 
sense, the unit of weight, the pound, may he taken as 
also the unit of mass (see articles 71, 72). 

But the weight Is also used as a measure of the force 
of gravity and of other forces compared with it. Hence 



56.] MASS— DENSITY — VOLUME. 53 

it must be carefully noted here that the term weight is 
employed with two distinct meanings, which should not 
be confounded; namely — 

(a) As a measure of the mass or quantity of matter in 
a given body. 

(b) As a measure of force by reference to the force of 
gravity. 

55. Distinction between Mass and Weight. Although 
the weight of a body may properly stand for its mass, if 
their true relation is understood, the two terms are not 
identical. The mass or quantity of matter of a lead 
ball is the same wherever it is situated on the earth's 
surface; but the weight, which may be registered on a 
spring-balance, is slightly greater at the poles than at 
the equator. Again, at the surface of the sun the force 
of attraction on the same piece of lead, registered as 
before by the stretching of a spring, would be about 
twenty-eight times greater than on the earth. Still 
further, if we conceive of it as at a point in space far 
away from attracting bodies, there would be no sensible 
pull on the spring, nothing to correspond to the terrestrial 
weight, but the mass would be everywhere the same. 

56. Relation between Mass, Density, and Volume* 

The density of a body is the mass or quantity of matter in 
the unit of volume. In comparing different bodies the 
density of water at the temperature of 39.2° F. (4° C.) is 
generally taken as unity ; the fact that the weight — that 
is, the mass — of a given volume of lead is 11-J times, or 
of iron 7 times, that of the same volume of water is 
expressed by saying that the density of lead is 11-J, and 
of iron is 7. A body is said to be throughout of uni- 
form density when equal volumes, however small, have 
the same mass. 



54 DYNAMICS. [57. 

In Art. 53 it was stated that for bodies of uniform 
density the mass is proportional to the volume. It also 
follows that for bodies of equal volume the mass is pro- 
portional to the density. 

Therefore, in general, the mass (M ) is proportional 
to the product of the volume (V) and density (D). 
This may be expressed mathematically in this form: 

M oc D V-, that is, -^ = ^y,; 
whence 

D a — , and V a -^-. 

57. Momentum. The momentum of a body is equal to 
the product of the mass and velocity. Two bodies of 
the same mass, and moving with the same velocity, have 
obviously the same momentum. If these two bodies 
were joined together, still retaining the same velocity -as 
before, the momentum of the two together as a whole 
would be twice that of either of them separately. In 
general, of two bodies having the same velocity, if the 
mass of one is five times that of the other, its momentum 
will be also five times as great; or, 

If the velocity is constant, the momentum is propor- 
tional to the mass. 

Again, suppose two bodies of the same mass, but one 
moving with twice the velocity of the other, its momen- 
tum will be also twice as great; or, in general, 

If the mass is constant, the momentum is proportional 
to the velocity. 

The unit of momentum is the momentum of a body 
of unit mass, moving with the unit velocity of one foot per 
second. A body whose mass is M and whose velocity is 



FORCE DEFINED. 55 

v has a momentum equal to the product of the mass 
into the velocity, or 

Momentum = Mv. 



EXAMPLES. 

XI. Mass — Density — Volume. Article 56. 

1. The masses of two bodies are as 2 to 7, and their densities as 
6 to 5: What is the ratio of their volumes ? 

2. The masses of two bodies are as 5 to 6, their volumes as 2 to 
3 : What is the ratio of their densities ? 

3. Two bodies of the same mass have densities as 8 to 9: What 
is their ratio in volume ? 

4. What is the ratio in volume of a piece of silver weighing 
20 lbs. and having a density of 10.5 (referred to water as unity), 
and a piece of iron weighing 5 lbs. and having a density of 7 ? 

5. If a cubic foot of water weighs 62.5 lbs. (density unity), what 
is the weight of a cubic inch of mercury, density 13.6 ? 

6. What is the ratio in weight (that is , in mass) of two blocks of 
stone, one having a volume of 50 cubic feet and a density of 3, the 
other a volume of 45 cubic feet and a density of 2.75 ? 

7. If a liter (1000 cubic centimeters) of water weighs a kilogram 
(density unity, temperature 4 s C. = 39.2° F.), and a cubic centi- 
meter of another liquid weighs 1.1 grams, at the same tempera- 
ture, what is the density of the latter liquid ? 

8. If a cubic foot of fresh water weighs 62.5 lbs., and of salt 
water 64 lbs. , what is the density of the salt water ? 

Kinds of Forces — Force of Gravity, 

58. Definition of Force. A force is that which moves 
or tends to move a tody, or which changes or tends to 
change its motion, either in direction or quantity. 

In view of the fact, before explained (11), that all 
bodies of which we have any knowledge are in motion, 
the completeness of the definition would not be im- 



56 DYNAMICS. [59. 

paired by the omission of the first clause. It is, how- 
ever, convenient to consider the earth and all bodies 
which do not change their position with reference to it 
as at rest, and the definition conforms to that idea. 
Further, the word "tend" is added because the action 
of one force may be neutralized by that of one or more 
opposing forces, so that the motion which it tends to 
produce is not observed. For example, a book resting 
on a table tends to fall to the ground under the action 
of the force of gravity, but an equal opposite force, the 
resistance or reaction of the table, keeps it at rest. 

59. Continued and Impulsive Forces. A force is said 
to be continued when its action continues an appreciable 
length of time. It is uniformly continued, or con- 
stant, when its intensity is always the same; this is true 
of the force of gravity at a given point on the earth's 
surface. A continued force is variable when its in- 
tensity is different at different times, as the force of a 
watch-spring, whose intensity diminishes as the spring 
unwinds. 

An impulsive force is one which acts through so short 
a time that the law of its action cannot be determined, 
and we are limited to considering its effects after its 
action has ceased. This is true of the blow from a bat 
on a ball. Such a force, however, is not strictly in- 
stantaneous, but one whose intensity is very great and 
varies during the brief time of its action. There is con- 
sequently no essential difference between the two classes 
of forces. 

60. Effects of Force upon a Free Body. A continued 
force tends to produce accelerated motion in the body 
acted upon. If the force is uniform as well, it tends to 



61.] FOKCE DEFINED. 57 

give the body uniformly accelerated motion. This is 
practically the motion of a body falling toward the earth 
under the influence of the nearly constant force of 
gravity (but see Art. 64). Therefore, if the accelera- 
tion produced by a constant force is / (for gravity g), 
the relations between the velocity acquired (v) and 
space passed over (s) in a given time (t) are expressed 
by the familiar equations (from Art. 27): 



v=ft, 


For gravity, v = gt, 


8 = iff, 


s = \gt\ 


v* = 2fs. 


v 2 = 2gs. 



The motion of a body acted upon by an impulsive 
force tends, after this force has ceased acting, to be 
uniform, as a ball struck along the ground by a bat. 
This is true, indeed, of any body in motion, and, as 
explained in Art. 67, is a consequence of the first law 
of motion. 

The presence of other opposing forces may modify the 
effect of the force considered. For example, a stone 
thrown vertically upward, and which consequently tends 
to move uniformly in that direction, has in fact re- 
tarded motion because of the continued and simul- 
taneous action of the force of gravity downward. So, 
too, a ball rolled along the ground, as a matter of ex- 
perience, soon comes to a state of rest because of the 
opposing force of friction. 

61. Equilibrium. A body is said to be in equilibrium, 
with respect to two or more forces, when they neutralize 
each other so that its condition of rest or motion is not 
affected by them. The book mentioned in Art. 58 is 
an example of equilibrium. But equilibrium does not 



58 DYNAMICS. [62. 

necessarily imply the rest of the body in question. For 
example, a ball rolling on a perfectly smooth horizontal 
surface is in equilibrium with respect to the two equal 
and opposite forces — the action of gravity and the re- 
action of the surface. The same would hold true how- 
ever many forces were involved if they, taken together, 
did not affect the motion of the body. Equilibrium 
strictly implies simply absence of acceleration. 

62. Examples of Forces. The first and simplest con- 
ception of force we derive from muscular exertion, as 
we note its effects in different ways. With it we join 
all other agencies which produce similar effects in 
changing the motion of bodies, as gravity, cohesion, 
electrical attraction and repulsion, and so on. 

63. Force of Gravity. The force of gravity is mani- 
fested in the attraction which the earth exerts on a mass 
of matter near its surface, and which causes it to fall, or 
«;end to fall, toward it. This is a special case .of the 
universal law of gravitation, established by Newton, and 
.according to which 

Every particle of matter attracts and is attracted by 
every other particle with a force which varies directly as 
the product of the masses and inversely as the square of 

MM' 

the distance. F a — 7r -. 
d 

A stone, therefore, as truly attracts the earth as it is 

attracted by it; so, also, the earth attracts and is attracted 

by the moon, the sun, and the other bodies of the solar 

system. In terrestrial mechanics, however, we have tc 

do simply with the attraction of the earth upon bodies 

on or near it, and as its mass is indefinitely great in 

comparison, the reciprocal attraction is left out of 



64.] FORCE OF GRAVITY. 59 

account. Therefore, since the mass of the earth is con- 
stant, the force of its attraction on any body varies 
directly as the mass of that body and inversely as the 
square of its distance from the earth's centre ; that is, 

M 

F oc -^. From this, it follows that for a given body 
a 

the force of attraction varies inversely as the square of 
the distance ; and for different bodies at the same dis- 
tance the force is directly as their masses. These two 
points are expanded in articles 64 and 65. 

The attraction between different bodies, although a force of 
small intensity where their masses are small, may be demon- 
strated in other ways than by the fall of a body to the earth. 
It is illustrated by the deviation from the usual perpendicular 
position, which is observed when a ball hung by a string is 
suspended near an isolated mountain. Experimenting in this 
way, Dr. Maskelyne found the angle of deviation for two plumb- 
lines, placed on opposite sides, north and south, of Mt. Schehal- 
lien in Scotland, and at a distance of 4000 feet from each other, 
to be 12 seconds. From this result the mean density of the earth 
was calculated to be about 5 times that of water. 

This attraction has also been shown by Cavendish in a more 
delicate manner, by means of the torsion balance. Two small 
balls of lead were attached to the ends of a slender wooden rod 
which was supported at the centre by a fine wire of considerable 
length. Two larger balls were then approached to the small ones 
and on opposite sides, so that their effect was felt in the same 
direction. The result was that the small balls were attracted by 
the larger ones, and the rod supporting them deflected from its 
original position of rest. The angle of deflection showed the 
amount of the torsion (or twist) of the wire, and this measured the 
intensity of the attracting forces. By comparing the attractions 
of these balls with that of the earth, Cavendish calculated the 
mean density of the earth to be 5.45. 

64. (1) The force of attraction of the earth on a given 
body varies inversely with the square of the distance. 



60 DYNAMICS. £64. 

F <x -;?. This law means that if the distance from the 
a 

attracting body be increased two, three, or ten times, the 

force of attraction it exerts on another body is i, ^, or 

Y^-g- respectively; if the distance is diminished to -J, -J-, 

and so on, the force of attraction becomes 4 times, 9 

times, etc., greater. 

Two consequences, of importance here, follow from 
this principle: 

(a) The attraction of the earth is sensibly constant at 
a given locality and for the different heights above the 
surface involved in ordinary observations. 

This attraction may be proved to be exerted as if the 
whole mass were concentrated at a point at or near the 
centre, called the centre of gravity. If now we call the 
radius of the earth in round numbers 4000 miles, the dis- 
tances from this centre for two bodies — the one at the sea- 
level and the other a mile above — will be 4000 and 4001 
miles respectively, and the ratio of the forces of attraction 
will be, as above, 400 1* : 4000 ; but this difference is so 
small that it may often be left out of account. In 
accurate physical hivestigations, however, this difference 
can by no means be neglected. Indeed, it should be 
most carefully noted that while gravity is here said to 
be sensibly constant, it really varies uninterruptedly as 
the distance from the centre increases, and is not abso- 
lutely the same for two points, one of which is a foot 
above the other. 

If we imagine a body to pass from the surface toward 
the centre of the earth, it may be demonstrated that the 
attraction will diminish, and if the earth were homo- 
geneous, in the same ratio as the distance from the cen- 
tre diminishes; at the centre the attraction is zero. 



65.] FORCE OF GRAVITY. 61 

(b) The force of attraction is least at the equator and 
increases toward the poles. 

The earth having the shape of an oblate spheroid, the 
polar diameter is about 26 miles shorter than the equa- 
torial, and hence, as it may be demonstrated, the force 
of attraction is greater for a body at the poles by T \j. 

To this cause for the variation of the intensity of the 
force of gravity is to be added a second, the rapid rota- 
tion of the earth on its axis. The effect of this cause to 
diminish the gravitation is greatest at the equator, and 
grows less as we go from it, and becomes zero at the 
poles (80). On this account, then, the attraction is 
greater by ^ at the poles than at the equator. As the 
result of the two causes taken together, the force of 
gravity at the poles is about yj-g- greater than at the 
equator. Thus the weight of a given mass of matter, 
by which the pull of the earth may be measured, in- 
creases as we go from the equator northward or south- 
ward toward the poles. This difference could be noted, 
for example, by observations with a sufficiently delicate 
spring-balance, and would amount to about 1 lb. in 
200 lbs. This difference is left out of account in ordi- 
nary commercial transactions, but cannot be neglected 
in physical problems where accuracy is required. 

The force of gravity also varies somewhat for differ- 
ent points on the earth's surface in consequence of varia- 
tions in the density of the material of the earth. This 
subject is expanded in a later article (250), where the 
values of g for different points are also given. 

65. Again: (2) The attraction of the earth on bodies 
at the same distance is proportional to their masses. 
F cc M. Hence the acceleration given to a falling body 
by the earth's attraction is independent of its mass. 



62 DYNAMICS. [65. 

For example, two pieces of lead of widely differing 
weights, or a piece of lead and a feather, fall the same 
distance toward the earth in the same time and gain the 
same velocity (that is, 32 feet per second for each second). 
But the experiment succeeds only when all disturbing 
causes — e.g. , the resistance of the air — are removed, so 
that an exhausted receiver must be employed. As a 
matter of experience, in comparing the fall of a small 
and heavy body and of a larger and lighter one, the 
former will fall the faster under ordinary conditions, 
but this is only because it feels less the resistance of 
the air. 

The fact here mentioned is based upon the above 
law, and is what a simple consideration would lead 
us to expect. Suppose several small 
• • • • • i shot as those at a (Fig. 28); it is 

Fig. 28. obvious that under the earth's at- 

traction they would fall together, keeping their relative 
position with reference to each other, and reaching the 
ground at the same time and with the same velocity. 
If now we suppose them rigidly connected, but their 
positions unchanged, they would form a mass 5 times 
greater than a single one (as 5), and attracted by a force 
also 5 times greater; they would still fall together, and 
the acceleration of this mass and the single one would 
be the same. This course of reasoning might be ex- 
tended to any two bodies, however unlike in mass. The 
force of attraction increases always in the same ratio 
that the number of particles, or the mass, of the body 
attracted increases, and hence the effect of the force of 
attraction must remain constant. 



66.] LAWS OF MOTION. 63 

EXAMPLES. 
XII. Force of Gravity. Articles 63, 64, 65. 

1. At what distance from the centre of the earth would a mass 
of matter weighing 32 lbs. on the earth's surface exert a pull 
equivalent to 1 lb. on a spring-balance ? 

2. If the mass of the sun is 350,000 times that of the earth, and 
its diameter 112 times, what is the acceleration of gravity at its 
surface ? (see also p. 67) 

3. If the moon's mass is ^ of that of the earth, and its diameter 
2160 miles, that of the earth being about 7900 miles, what is the 
acceleration of gravity on the moon's surface ? 

4. If the acceleration of gravity on the surface of Jupiter is 
2.62 times g, and its diameter 11 times that of the earth, what is 
their ratio in mass ? 

5. If the value of g at the equator, at the sea-level, is 32.096, 
what is its value at the summit of a mountain in latitude 0°, at an 
altitude of 15,840 feet ? The equatorial radius is 3963.3 miles. 

Newton's Laws of Motion. 

66. Laws of Motion. The three laws of motion, as 
stated by Newton, are: 

(1) Every body continues in a state of rest, or of uni- 
form motion in a straight line, except in so far as it may 
be compelled by impressed forces to change that state. 

(2) Change of motion is proportional to the impressed 
force, and takes place in the direction of the straight 
line in which the force acts. 

(3) To every action there is always an equal and con- 
trary reaction; or, the mutual actions of two bodies are 
always equal and opposite in direction. 

The truth of these laws is established by observation 
and experiment; it is found that all legitimate conclu- 



64 DYNAMICS. [67. 

sions deduced from them are in harmony with the ob- 
served facts of nature. 

67. The first law asserts what is sometimes called 
the inertia of matter; in other words, that matter alone 
is powerless to change its state either of rest or motion. 
These points may be considered separately. 

(a) The tendency of a body at rest to remain in this 
condition is universally recognized; it is always mani- 
fested by the apparent resistance which such a body 
offers to a force tending to set it in motion. This 
apparent resistance of heavy bodies has nothing to do 
with that caused by other opposing forces; e.g., friction. 
It increases, as stated in 53, with the mass of a body, 
and, as explained under the second law of motion (68), is 
due solely to the fact that in such cases a force must 
continue to act through a certain time in order to im- 
part sensible motion. For example, suppose a heavy 
fly-wheel of an engine free to turn on its axis without 
friction, or a heavy cannon-ball suspended by a string 
of great length, or a massive iron door well poised on 
its hinges; in all such cases the hand in trying to move 
the bodies seems to encounter resistance, which ex- 
presses the inertia of the body, or its tendency to 
remain at rest. The only real difference, however, 
between the bodies named and a light body moved with 
a touch arises from the greater mass involved in the 
former case; both alike have inertia. 

This inertia, or resistance to motion, is a property of 
all forms of matter ; it is manifested by the water when 
a boat moves rapidly through it, and by the air when 
one is driving or running rapidly. 

(b) The application of the second part of the law is 
equally familiar, although it is impossible to give an 



67.] LAWS OF MOTION. 65 

experimental proof of its truth. It requires that a 
body once in motion shall — unless acted upon by some 
force tending to change its motion — continue to move 
forever uniformly and in a straight line. This tendency 
toward continuance in a state of motion is seen in the 
apparent resistance that bodies in motion make to a 
force tending to stop them. We observe, also, that if a 
carriage in rapid motion is suddenly stopped, the per- 
sons occupying it are thrown violently forward; so, too, 
if a person steps off from a train in rapid motion, his 
body tends to keep its forward motion, and when the 
motion of the feet is arrested by touching the ground, 
the rest of the body is thrown forward and a fall is the 
result. 

But a body in motion not only tends to move on uni- 
formly, but also in a straight line. Hence if we observe 
a body moving in a circular path, as a stone attached to 
a string and whirled about a centre, we are justified in 
concluding that a force is continually acting to deflect it 
from a straight line. If the string to which the stone is 
attached breaks, the stone flies off at a tangent to its 
former course. So, too, of the earth: it moves about 
the sun in a path nearly circular, its tendency to move 
in a straight line being overcome by the continued 
attraction of the sun. 

The tendency to uniform motion in a right line is all 
that can be observed; for, as stated above (60), uniform 
motion in a body not acted upon by any force is alto- 
gether contrary to experience — any body, once set in 
motion, sooner or later comes to rest. But in every such 
case it is possible to trace the more or less rapid loss of 
motion to outside causes; the most universally present 
are friction, or the resistance to motion due to the rough- 



DYNAMICS. [68. 

of the surfaces in contact, and the resistance of 
the air, which last is an important element in the case 
of rapid motion, as that of a bullet. 

The truth of the law is argued from the observed fact 
that, in proportion as these opposing forces are removed, 
the motion continues longer and longer. A ball, which 
with a given impulse will roll a certain distance on a 
horizontal surface of turf, rolls farther on gravel, farther 
still on a marble floor, and still farther on a sheet of ice. 
So, too, the time which a pendulum, once set in motion, 
will continue to vibrate without additional impulse 
becomes longer and longer as we remove the friction on 
its axis of support, and the resistance of the air by 
placing it in an exhausted receiver. The case of the 
earth is the most perfect illustration of this law, for it 
moves on in its orbit with a mean velocity that the most 
accurate observations can hardly prove to vary, although 
it is receiving no forward impulse; the sun's attraction, 
as explained above, only serves to keep it in its nearly 
circular orbit. 

68. The second law of motion asserts (a) that the 
change of motion is proportional to the impressed force. 

By motion is meant here momentum, or the pro- 
duct of the mass and velocity, as defined in Art. 57. 
The law consequently asserts that the change of mo- 
mentum in a given time is proportional to the force 
which acts, and hence, as explained in 71, this change 
of momentum is a measure of the force. 

Let F represent the force, M the mass of the body 
moved, and / the velocity given to it in one second 
(the acceleration). Then Mf is the momentum gener- 
ated in one second, and by this law F is proportional to 
Mf ; that is, F is equal to Mf multiplied by some con- 



68.] LAWS OF MOTION. 67 

stant number; if suitable units are taken, this constant 
becomes unity, and then 

F=Mf. (1) 

If the force acts for t seconds with uniform intensity, 
then its effect will be proportional to the time (= Ft), 
and the velocity given to the body at the end of this 
time will be ft or v. Hence, with the same provision 
as above, we have 

Ft = Mv. (2) 

In the case of gravity the momentum generated in a 
given time — that is, in one second — is Mg. But since 
the weight of the body, expressed in standard pounds, is 
proportional to the mass (M), and also to the accele- 
ration of gravity (g), the weight is also proportional to 
the product Mg; if a suitable unit of mass is taken, we 
may write 

W=Mg, (3) 

or 

W 
M = —. (4) 

The expression in equation (4) is the value of the mass 
in terms of the weight in pounds which is ordinarily 
employed in Mechanics. 

From equation (2) the following principles are de- 
duced: 

1. The velocities given in the same time to different 
todies of the same mass are proportional to the acting 
forces. That is, if forces whose intensities are as 1,2, 3, 
act on three bodies of equal mass, the velocities gene- 
rated in the same time will be in the ratio of 1 : 2 : 3. 
For example, the intensity of the sun's attraction at its 
surface is about 28 times, and of Jupiter 2.6 times, that 



68 DYNAMICS. [68. 

of the earth; therefore the velocity acquired at the end 
of one second by a falling body, on the sun, on Jupiter, 
and on the earth, will be respectively 28 X 32 feet per 
second, 2.6 X 32, and 32. 

2. If equal forces act upon bodies of different mass 
for the same time, the velocities will be inversely propor- 
tional to the masses. A force which would give a body 
of mass 1 a velocity of 12 feet per second in a certain 
time, would give a body of mass 3 a velocity of only 
4 feet per second in the same time. This principle 
shows the reason why a "heavy" body — that is, one of 
great mass — seems to offer more resistance to motion 
than a lighter one. For suppose the ratio in mass to 
be 10 : 1; then the same force acting upon the heavier 
body will give it in the same time the same momentum, 
but only -fa the velocity. In other words, it must act 
through ten times the length of time in order to gene- 
rate the same velocity. 

Again — 3. To give bodies of different mass the same 
velocity in the same time, the forces must in each case be 
proportional to the masses. As explained in 65, this is 
true of the force of the earth's attraction, which gives to 
all falling bodies at its surface the same acceleration. 

Finally — 4. // equal forces act upon bodies of equal 
mass, the velocities generated will be proportional to the 
times of action. 

(b) The second law also states that the direction of 
motion is that of the impressed force. When a force 
acts upon a material particle at rest — that is, a portion 
of matter so small that its dimensions may be left out of 
account — the truth of this law is evident. If a body is 
acted upon, it will in general take the direction of the 
impressed force only when the line of its action passes 



69.] LAWS OF MOTION. 69 

through a point in it called the centre of inertia. When 
this is not the case, as when a ball is struck upon its 
side by a bat, there is a tendency to rotate, and the body 
moves in a direction which varies more or less from that 
of the force. 

In general, if any number of forces act upon a body, 
this law will hold good for each of them as though the 
others did not exist ; or the effect upon the body, as re- 
gards quantity and direction of motion, at the end of any 
time, is the same as if the forces had acted each through 
the same time in succession. 

So far as it relates to the direction and rate of motion 
this is properly the basis of the Parallelogram of Veloci- 
ties, explained in Art. 33. For example, in the cross- 
ing of the stream by the uniformly moving boat (3G), 
each motion goes on independently, not modified by the 
existence of the other, although the position of the boat 
at any moment depends on the action of both. Again, 
every motion goes on aboard a steamboat indifferently 
whether the boat is or is not in motion : a ball thrown 
up by a person standing on deck falls again to his hand, 
and if dropped from the top of the mast falls at its foot 
a second or two after, although the motion of the boat 
may have been rapidly forward all the time. So, too, a 
rifle-ball fired horizontally from a window, and another 
dropped vertically down at the same instant, reach the 
ground in the same time (allowance being made for the 
resistance of the air). In each case the forward and 
the downward motions go on independently of each 
other. 

69. The third law is sometimes briefly stated in this 
form, that: The action and reaction are equal and con- 
trary. 



70 DYNAMICS. [70. 

The general term stress is employed to express the mutual 
action between two bodies, for the action of a force always 
implies two bodies, and this law affirms that the action of the one 
is exactly equal to the reaction of the other. 

This law is true whether the force is exerted as pres- 
sure, as a pull or tension, or as a blow. 

(a) Pressure, For instance, when the hand presses 
against the wall, a contrary and equal reaction is exerted 
by the wall. A weight, exerting pressure on a support, 
encounters an equal return pressure upward. 

(b) Tension. For example, a ten-pound weight, hang- 
ing by a string, exerts a pull of 10 lbs. , and the reaction 
of the hook to which the string is attached is also 10 lbs. 
Also, if a horse drags a heavy load by a rope, the load 
pulls back an amount equal to that which the horse 
exerts. 

(c) Bloio. When a hammer strikes a nail, the reaction 
is equal and contrary. Again, when a cannon is fired 
off, the recoil of the gun and carriage is due to the 
reaction equal to the action in driving forward the shot. 

The third law of motion, taken in its broadest sense, 
is true only when the apparent loss of mechanical energy 
on impact is allowed for, as explained later (108 et seq.). 

70. Collision of Inelastic Bodies. It follows from the 
preceding article that: When several bodies come into 
direct collision, the momentum of the whole system before 
and after impact is the same. 

Suppose two inelastic bodies whose masses are M and 
m, and whose velocities are V and v; the momentum of 
the first is M V, and of the second is mv. (a) If they 
are moving in the same direction, the momentum of the 
two is 

MV-\-mv. 



70.] LAWS OF MOTION. 71 

If now the faster-moving body overtakes and impinges 
upon the other, the two after impact will move along 
together with a velocity v' less than V and greater than 
v; the momentum of the two together will be 

(M + m) v', 
and by this law 

M V + mv = (M + m) v\ (1) 

(b) If the two bodies were moving in opposite direc- 
tions, then the momentum of the two moving separately 
is 

MV — mv, 

and after impact of the two together, 

(M -f- m) v'\ 
and, as before, 

MY — mv = (M+ m) v'. (2) 

As an example of this, suppose that a rifle-ball weigh- 
ing one ounce and moving with an unknown velocity v 
strikes and penetrates a body whose weight is 10 lbs., 
and that after impact the velocity of the mass of wood 
with the imbedded ball is v' = 8 feet per second; then 
^ X v = (10 + tV) X 8. 

From which it may be calculated that v — 1288. 
This is the principle involved in the ballistic pendulum; 
the velocity after the impact is determined, however, 
not directly but by calculation from the height to 
which the mass is raised (v 2 = 2gh). 

If the impinging bodies are perfectly or imperfectly 
elastic, the conditions are changed, and a factor express- 
ing the degree of elasticity (coefficient of elasticity) 
must be introduced. The discussion of these cases lies 
outside of the scope of the present work. 



72 DYNAMICS. [71. 

EXAMPLES. 

XIII. Collision of Inelastic Bodies. Article 70. 

[The bodies are supposed to be perfectly inelastic, and their mo- 
tion is uniform ; the impact is direct, not oblique. ] 

1. A ball weighing 10 lbs. and having a velocity of 16 feet 
per second overtakes a second ball weighing 5 lbs. and whose 
velocity is 8 feet per second : What is the final velocity ? 

2. If the first ball in the preceding example meets the second, 
what is the final velocity ? 

3. A body weighing 40 lbs. strikes another at rest weighing 360 
lbs. , and the two move on with a velocity of 2 feet per second ; 
What was the original velocity of the first ball ? 

4. Three bodies, each weighing 4 lbs. , are situated in a straight 
line ; a fourth, weighing 8 lbs. and moving at a rate of 12 feet per 
second, strikes them in succession: What velocity results after 
each impact ? 

5. Two bodies moving in the same direction at the rates of 8 
and 10 feet per second come into collision, and after impact have 
a velocity of 8.4 feet per second: What is the ratio of the masses 
of the two bodies ? 

6. If the bodies in example 5 move in opposite direction, and 
the final velocity is .4 feet per second, what is the ratio of their 
masses ? 

7. A body moving 10 feet per second meets another moving 
2 feet per second, and thus loses one half of its momentum : What 
is the ratio of the masses of the two bodies ? 

8. A body weighing 6 lbs. strikes another weighing 5 lbs. and 
moving in the same direction at a rate of 7 feet per second : If the 
velocity of the second body is doubled by the impact, what was 
the previous velocity of the first body? 

9. An ounce rifle-bullet is fired (as in 70) into a suspended block 
weighing 36 lbs. ; the blow causes the wood to rise 1^ inches: Re 
quired the velocity of the bullet at the moment of impact. 

Measurement of Force. 
71 Absolute Method of Measuring Force. A force 
may be measured: By the velocity which it gives the unit 



72.] MEASUKEMENT OF FOKCE. 73 

of mass in the unit of time. The unit force is then a 
force which will giye a pound of matter a velocity of one 
foot per second in a second. This unit is sometimes 
called a poundal. It may also be stated in this equiva- 
lent form, already implied in Art. 68: A unit force is 
one which will generate (or destroy) a unit of momen- 
tum in one second. 

When the units of the metric system are employed, a unit force 
is defined as one which will give one gram of matter a velocity of 
one centimeter per second in one second ; this unit force is called 
a Dyne. 13,825.38 dynes make one poundal. This system of 
measuring force is called the centimeter-gram-second system, oi 
the C.G.S. system. 

The force of gravity on a pound of matter, which giveb 
a velocity of about 32 feet per second in a second (g), is 
then a force of 32 poundals, and hence this number 32 
(or g) is the measure of the earth's attraction on this 
absolute system. As it is 32 times the force required to 
give one pound a velocity of one foot per second, it is 
evident that the unit force, the poundal, is equivalent 

to the action of gravity on about half an ounce \—^r 

\ OX) 

This method, of measuring force is called absolute 
measure, in the sense that it is universally applicable and 
independent, as the following method is not, of the vari- 
ations in the force of gravity. As implied above, the 
unit of mass in this system is the standard pound. 

72. Gravitation Method of Measuring Force. Forces 
are also measured: By comparing them directly with 
gravity; that is, by the weights they could support. The 
unit fokce is then a force equal to that required 
to support the standard pound against the force of 
gravity; or, briefly, it is equal to the weight of one 



74 DYNAMICS. [72. 

pound. It is then g times, or about 32 times, the unit 
of force mentioned in the preceding article. This is 
called gravitation measure. 

Since now the force of gravity manifests itself every- 
where and at all times on the earth, and since we are 
so familiar with its intensity as measured by the weights 
of one, two, ten pounds, and the amount of muscular 
exertion required to overcome it, this is a most simple 
and natural way of measuring all forces. After this 
method we say that the tension of the rope pulling a 
canal-boat is 100 lbs. when the force exerted is equal to 
that required to support a weight of 100 lbs. It is 
common to speak of a pull — as on an oar — of 50 lbs., of 
the force of the wind or that of the waves as being so 
many pounds, etc. In cases like the last a dynamometer 
is employed, and the pressure on a spring noted, and this 
readily compared with the same effect produced by a 
known weight under the action of gravity. 

Notwithstanding the fact that this method is so com- 
monly and conveniently employed, it is less scientific 
than the absolute method, and is open to one serious 
objection, that it does not necessarily take into account 
the variations in the force of gravity. As explained in 
Art. 64, the force of gravity varies about -^ between 
the equator and the poles, and, when the same mass is 
used as the unit of weight, a pull of a pound means a 
stronger pull in high latitudes than toward the equator. 
Hence the gravitation method is accurate only when the 
difference in the value of g at the spot in question and 
at the sea-level is known and taken into account. 

When forces are considered as producing accelerated 
motion, the absolute measure is generally employed; but 
when they act, as usually in Statics, either as tensions or 






73.] DYNAMICAL PEOBLEMS. 75 

pressures, the gravitation measure is the one usually 
accepted. 

The u:ntt of mass employed (in the gravitation meas- 
ure of force) is the quantity of matter in a body which 
weighs g pounds, where g is the acceleration of gravity 
for the place in question. This is then a varying unit, 
having a definite value for each place under considera- 
tion. The reason for selecting this unit is that the 
numerical expression of the mass of a given body will be 
the same everywhere; that is, a body of mass 10 will still 
be this wherever it is. For since, as has been shown (68), 

W . W 

we may take M = — , therefore this ratio of — , and 

hence the numerical value of M, will be always the same. 
For example, on the sun, where the force of attraction is 
about 28 times that on the earth, we should have 
XT 28. TF 

or the same value as before. 

Problems in Dynamics. 

73. In Art. 68 it was shown that the intensity of any 
acting force (F) is equal to the product of the mass 
moved (M ) into the velocity generated in one second; 
that is, 

F = Mf. (1) 

If in this equation we substitute the value of M I =— - J, 

W 
we obtain F =--./, (2) 



76 



DYNAMICS. 



[74. 



This relation makes it possible to obtain the accelera- 
tion (/) produced by any known force acting upon a 

body whose weight is also 
known. When / is known, 
then the equations of articles 
27, 41, 42 give the means of 
calculating all the particulars 
in regard to the motion of the 
body; that is, the space passed 
through in a given time, the 
velocity acquired, etc. 

74. Attwood's Machine. 

Equation (3) in the preced- 
ing article, in connection 
with Attwood's machine, 
makes it possible to verify 
the laws of motion given in 
articles 67, 68. The essen- 
tial parts of the machine are 
shown in Fig. 29. There is 
a pulley over which a thread 
passes, holding any weights 
P and Q. The axis of the 
pulley rests on two smaller 
wheels, called friction-ivheels, 
which serve to diminish the 
friction so that its effect can be 
disregarded. The whole is sup- 
ported by a firm, massive stand. 
In the path of the weights 
may be clamped a stage (as d) 
at any point, as determined by 
the vertical scale/). A ring (c) 




Fig. 29. 



74.] DYNAMICAL PEOBLEMS. 77 

in the same position serves, if desired, to remove a por- 
tion of the weight without disturbing the motion of the 
rest. For this last purpose the extra weight is in the 
form of a straight bar (b), which is caught by the ring 
as the weight moves through it. A pendulum to beat 
seconds is added to the whole. 

The following are some of the experiments which may 
be tried: 

(1) Suppose the weight P = 8 J oz. and Q = 7f oz., 
supported on either thread; then the weight moved ( W) 
is = P + Q = 16 oz. (= 1 lb.), and the moving force (F) 
is the difference, or P — Q = ^ oz. [Strictly the weight 
moved also includes the weight of the pulley, which in 
an accurate experiment would have to be taken into 
account.] Therefore, by equation (3), if g = 32, 

f= -j^r. g = p *X .ff = 1 f oot-per-second per second. 

Now, by Art. 27, the space passed through in the 
first two seconds (t = 2) by a body moving with uni- 
formly accelerated motion is 2f. Clamp the stage at a 
distance of 2 feet below the initial point of P, and the 
weight descending will strike the stage exactly two 
seconds after starting. This is also a verification by 
experiment of the remark, made in Art. 71, that the 
unit force, the poundal — that is, the force which would 
give the unit of mass 1 lb. (here P + Q) in a second a 
velocity of one foot per second — was equal to the weight 
ofioz. (= P-Q). 

If the stage be clamped at 4^ feet, then the weight 
will strike it at the end of three seconds, as the formula 
s — \ff requires. Showing, too, that in three seconds 
the space passed through is f times that in two seconds; 



78 DYNAMICS. [74. 

or, in other words (27), the space is proportional to 
the square of the time (2 a : 3 2 = 4 : 9). 

(2) Again, let P = 8.5 oz. and Q = 7.5 oz.; then 
P + § == 16 oz., or the weight moved is the same as in 
(1), but the moving force P — Q = 1 oz., or twice that 
in (1); then, as before, 

P — Q 
f = p . ^ . # = 2 f eet-per-second per second. 

Hence, in accordance with the law stated in Art. 68, 
the mass remaining constant, the velocity generated in a 
given time is proportional to the force acting; that is, 

Z. L I 

F' f ~ 2' 

If, as before, the stage is clamped 4 feet below the 
starting-point, the weight will strike it at the end of two 
seconds, as the formula requires (s = iff). 

(3) LetP = 4ioz. and Q = 3foz.;thenP+<) = 8 oz.; 
that is, the weight (or mass) moved is one half that in 
(1), while P — Q = | oz. ; that is, the moving force is 
the same. Equation (3) gives 

P — Q 

f = p ^ g — 2 f eet-per-second per second. 

That is, in accordance with the law stated in Art. 68, 
the acting force being constant, the velocity generated in 
a given time is inversely as the masses acted upon, 

f _ m_ J16 _ 2 
/ " = m 9 ~ 8 ~~ I' 

This value of/ may be verified as before. 

(4) Let P = 8.5 oz. and Q = 7.5 oz., and let the 
excess of P over Q (1 oz.) be in the form of a rod pro- 



75.] DYNAMICAL PEOBLEMS. 79 

jecting so as to be removed by the ring. Then, for this 
case, / = 2; hence if the ring be placed one foot below 
the starting-point, the weight will reach it at the end of 
one second. Here the extra weight is left behind, and 
now, the two weights being equal, the bodies must, 
according to the first law of motion, move on with uni- 
form velocity. At the end of another second it will 
strike the stage if placed 2 feet below — that is, 3 feet from 
the starting-point — and at the end of two seconds at 5, 
and so on. 

75. The following is a similar application of the 
above principle. Let W (Fig. w 

30) be a weight resting on a W A - 

perfectly smooth horizontal «& 

plane; P is another weight 

attached to W by a string 

passing over the pulley a; 

then, neglecting the weight Fig. 30. 

of the pulley, when P falls it moves W also; hence the 

total weight moved is P + W, and the moving force is 

P. Hence equation (3) becomes 

,_ _P_ „ 
1 ~ P + W' 9 ' 

The tension (T) of the string is given from the rela- 
tion 

I -f 

substituting in this the above value of/, we obtain 
T- PW 

P+W 



80 DYNAMICS. [7a 

76. The relation (3) in Art. 73 is applicable to a re- 
tarding force such as friction. For example, let W be a 
weight moving on a rough horizontal plane; the force of 
a friction (F) is a force acting parallel to the surface 
in a direction opposite to the motion. The retardation 
due to friction is given by the equation 

If F is known, and also W, then / can be calculated; 
this is the retardation due to friction; in other words, 
the body loses each second in velocity / feet per sec- 
ond. The distance which the body will go through 
before coming to rest, and its distance at any moment 
from the starting-point, will be given by the formulas 

V = u -ft, 
s = u t — ift\ 
Other illustrations are given in the examples below. 

XIV. General Dynamical Problems. Articles 68, 73-76. 

1. If in Attwood's machine P= 4£ oz. and Q = 3-$- oz. : (a) What 
is the acceleration? (b) What space will be passed through in 
2 seconds? 

2. (a) At what height above the earth's surface would a body- 
fall 4 feet in the first second from rest? (b) If its weight was 40 
lbs., what pull would it exert on a spring-balance at this point? 

3. A 12-lb. weight hanging over the edge of a smooth table 
drags a 60-lb. weight with it: What is the acceleration and the 
tension of the string? 

4. If the table in example 3 is rough, and the resistance of 
friction consequently equivalent to one tenth of the weight of the 
sliding body, what is the acceleration? 

5. (a) A bucket weighing 100 lbs. is raised up from a well at 



76.] DYNAMICAL PEOBLEMS. 81 

a uniform rate of 12 feet per second : What is the tension of the 
rope? (b) If the acceleration is 4 feet-per-second per second, 
what is the tension? 

6. For what time must a force of 4 oz. (gravitation measure) 
act on a body weighing 8 lbs. to give it a velocity of 20 feet per 
second? 

7. Two weights of 16 and 14 oz. hang over a pulley: What 
space will they move through from rest in 3 seconds? 

8. Two weights, each 8 oz., hang over a pulley: What addi- 
tional weight must be added to one of them to give an accelera- 
tion of 2 feet-per-second per second ? 

9. A weight of 8 lbs. rests on a smooth horizontal table 12 feet 
wide: What weight hanging vertically will draw it across in 
3 seconds? 

10. A weight of 24 lbs. rests on a platform : (a) What is its 
pressure on the platform if the latter is ascending with an accele- 
ration of i<7? (b) If descending with the same acceleration? 

11. A body weighing 160 lbs. is moved by a constant force, 
which generates a velocity of 8 feet per second : What weight 
could the force support? 

12. A force of 8 lbs. (gravitation measure) acts constantly on a 
body weighing 24 lbs. and resting on a smooth horizontal sur- 
face : (a) What is the acceleration, and (b) how far will the body 
move in 4 seconds? 

13. The velocity of a body weighing 24 oz. is increased from 
20 to 40 feet per second while the body passes over 30 feet : What 
is the moving force? 

14. Of two weights hanging over a pulley, one is 1 lb. and it 
ascends with an acceleration of 10 feet-per-second per second: 
What is the other weight? 

15. How long must a constant force of 10 lbs. act on a mass of 
100 lbs. to give it a velocity of 30 miles an hour? 

16. What constant force will cause a body weighing 400 lbs. to 
pass over 1200 feet in 10 seconds from rest on a smooth horizontal 
surface? 

17. A constant force of 15 lbs. gives a body an acceleration of 
5 feet per second in one second: What is the weight of the body? 



18. A body weighing 24 lbs. is projected on a rough horizontal 



82 DYNAMICS. [76. 

surface, where the resistance of friction is 6 lbs., with an initial 
velocity of 64 feet per second : (a) How far and (b) how long will 
it slide before coming to rest? 

19 A body weighing 40 lbs. is projected as in the preceding 
example. What is the resistance of friction if the body slides 
16 seconds before stopping? 

20 A body weighing 60 lbs. is projected up a rough plane in- 
clined at an angle of 30°, where the resistance of friction is 6 lbs., 
and with an initial velocity of 160 feet per second : (a) How far 
on the plane will it ascend? (b) How long will it take? 

21. The length of an inclined plane is 1000 feet, and its base 
800 feet; the resistance of friction is one eighth of the weight: 

(a) What initial velocity must it have just to reach the top, and 

(b) how long will the ascent take ? 



CHAPTER III— CENTRAL FORCES: 

77. Uniform Circular Motion. Suppose a body to be 
moving in a circular path with a constant velocity ; 
according to the first law of motion (66), a body in 
motion, if not acted upon by any force, tends to move 
on uniformly in a straight line. In order, therefore, 
that this body should, as supposed, move in a circle, it 
must be acted upon by a constant force exerted in the 
direction of the centre of the circle. 

This force toward the centre, or central force, is called 
the centripetal force. The equal and opposite (69) 
reaction exerted away from the centre is called the 
centrifugal force. The central forces determine the 
direction of motion of the body, but do not affect its 
rate of motion or velocity, since they act continually at 
right angles to its path. If a body attached to a string 
be whirled about a centre, the intensity of these cen- 
tral forces is measured by the tension of the string. 
If the string be cut, the body will fly off in a tangent to 
the curve, but with unchanged velocity. 

78. To find the intensity of the central force in the 
ca<e of uniform circular motion : Suppose the body to 
be moving at a constant velocity v about the circle ADH 
(Fig. 31), whose radius {AC) is r. In a very short space 
of time (t) it will have gone from A to D, so that 

AD = vt. (1) 



84 



DYNAMICS. 



[78. 



But in this time it has been drawn away from the tan- 
gent toward the centre a distance equal to BD (or AE). 
Let /be the acceleration of the constant central force; 
then (60) 

AE = iff. (2) 

But since AFH is a semicircle, the angle ADH is a right 
angle, and, by geometry, 



AD = AE X AH = AE X 2r. 



(3) 



If AD is taken very small, the chord AD may be 
regarded as identical with the arc AD. Therefore, 




introducing into (3) the values given in (1) and (2), we 

have 

v 2 f = iff X 2r, 



or 



/ 



(4) 



This value of / (4) gives the acceleration due to the 
constant force in terms of the velocity and the radius of 
the circle. From it we see that the value of / varies 



79.] CENTRAL FOKCES. 85 

directly with the square of the velocity, and inversely as 
the radius. 

In order to obtain the intensity of the force (F) — that 
is, in the case supposed above, the tension of the string — 
the value of/ must be multiplied, as explained in 68, by 
the mass (M) of the body in motion. "We have then 

F=Mf= m. 

If the value of F in pounds be required, when the 

W 
weight (W) is given: since M = — , we have, further, 

F=^X. 
9 r 

79. The pull away from the centre, called the cen- 
trifugal force, is felt whenever a body is made to rotate 
rapidly about a fixed centre. It is exemplified by the 
case of a loaded sling: if the cord is elastic, the extent 
to which it is stretched is a measure of this force. Simi- 
larly, a bucket containing water may be swung around 
by a rope so rapidly that this force becomes greater than 
that of gravity, and the contents are consequently not 
lost even when it is inverted. 

In the case of a large wheel rotating rapidly, if the 
centre of gravity and the axis of rotation coincide, the 
effects upon the parts on opposite sides of the axis neu- 
tralize each other and produce no result, except when 
the force becomes greater than the cohesion between 
the particles, when fracture takes place — as when a 
grindstone breaks. If, however, the centre of gravity 
does not coincide with the axis, a continuous pull on the 
bearing is produced by the motion which may lead to 



86 DYNAMICS. [80. 

very injurious results. For the same reason a crank-arm, 
which in use is turned rapidly, is generally weighted on 
the side of the centre opposite to the handle, to neutral- 
ize the injurious pull on the axis that would otherwise 
exist. 

It* a globe containing a little mercury be set in rapid 
rotation, the effect is to cause the mercury to recede 
from the axis and hence to rise and form a ring about 
the central part farthest from the axis. The governor 
of Watt, applied to the steam-engine, consists essen- 
tially of two heavy balls carried on rods jointed at the 
top. They are connected with some turning part of the 
engine, and, on the above principles, an increase in the 
rate of revolution causes them to separate and rise, and 
conversely if the rate is diminished. The arrangement 
is such that in the former case they partially close, and 
in the other case open, a valve by which the supply of 
steam is received. They thus serve to regulate the 
motion and keep it uniform; whence the name the 
instrument has received. 

80. Centrifugal Force due to the Earth's Rotation. 

Since the earth is rotating on its axis, every body on its 
surface, tending to move on in a straight line, must be 
retained there by a force pulling toward the axis. In 
other words, a certain portion of the earth's attraction 
for every body on its surface is exerted simply to con- 
strain the body to move in a circle, and is consequently 
not felt as weight. This is equivalent to saying that the 
centrifugal force acts on every body directly away from 
the centre of the circle in which it is moving. The di- 
rection of this force is indicated for the points E and B 
(Fig. 32) by the lines EG and BK. 



81]. 



CENTRAL FOltCES. 



87 



Since, by the preceding article, 

f= v l r W - r*o- 

the value of / can be calculated for the equator, for 
the value of v is given by the fact that any point on it 




describes a distance equal to the equatorial circumference 
in 24 hours. In this way we obtain 

/ = .1112 feet-per-seeond per second. 

This value of/ is about ^-^ of the value which g would 
have at the equator if this influence did not exist; since 
17 2 = 289, it follows that, if the velocity of rotation of 
the earth were increased 17 times, all bodies upon it 
at the equator would entirely lose their weight. 

81. For any point as B (Fig. 32) where the latitude 
is Z, the value of this force /', and the velocity v f (= v 
cos I), we have 



f = — = 



V COS I V , 

■ T = — COS I, 

r cos I r 



'./'-= fcosl 



88 DYNAMICS. [81. 

The components of f {BE, Fig. 31) are BH normal 
to the surface and BL as a tangent. Of these, since 
HBK=l, 

BH = fcosl=fcos?l; 

■»-»■»- /«*•■» /»•■» i t» sin /&i 
BL = f sm I = / sm I cos / = / — «- 

The normal component alone inflnences the weight of 
the body as it acts directly contrary to gravity, while the 
tangential component tends to produce motion toward 
the equator. It was the influence of the tangential com- 
ponent when the earth was in a plastic condition which 
is believed to have caused the flattening at the poles. 
This effect may be illustrated by the rapid rotation of a 
plastic mass of clay on its axis. 

The value of/ is greatest at the equator and dimin- 
ishes, with the cosine of the latitude, as we go toward 
the poles; at the poles it is zero (cos I = cos 90° = 0). 
The normal component is greatest at the equator and 
diminishes with the square of cosine of the latitude. 
The tangential component is zero at the equator, in- 
creases to latitude 45° iBL = ^) and diminishes from 
there to the poles, where it is again zero. 

EXAMPLES. 

XV. Centripetal and Centrifugal Forces. Articles 77-81. 

1. A ball weighing 20 lbs. is whirled by means of a string 
around a centre at a radius of 7 feet, with a linear velocity of 28 
feet per second- What is the value of/, and what is the tension of 
the string (F) ? 

2. (a) If the velocity is doubled in the preceding example, what 
do the values of/ and F become ? (b) What are they if the radius 
is doubled ? 



81.] CENTRAL FORCES. 89 

3 A ball weighing 4 lbs. attached to a centre at a distance of 8 
feet makes 300 revolutions in a minute : What is the pull on the 
centre ? 

4. What linear velocity of rotation must a body have if the 
tension of the string by which it is attached to the centre, at a 
distance of 8 feet, is equal to its weight ? 

5. What is the angular velocity of a body moving in a circle 
with a radius of 4 feet, when the centrifugal force is one half the 
weight ? 

6. A locomotive weighing 12 tons moves at a rate of 30 miles 
an hour about a curve whose radius is 1000 feet : What is the 
horizontal pressure on the rails ? 

7. If a stone weighing 5 lbs. is attached to a string 3 feet long 
and makes two revolutions in a second, what is the pull on the 
centre ? 

8. If a string can just support a weight of 400 lbs., what is the 
greatest length that can be employed to swing around a 20-lb. 
weight once in a second ? 

9. What is the shortest length of the string only strong enough 
to support 100 lbs. that can be used to whirl around a 50-lb. 
weight at a rate of 8 feet per second ? 



CHAPTER IV.— FRICTION. 

82. Definition of Friction. Friction is the resistance 
which is offered to the motion of one body upon another 
due to the roughness of the surfaces in contact. Fric- 
tion always acts parallel to the surfaces, and in a direc- 
tion contrary to that in which the body is moving or is 
about to move. 

83. Reaction of Smooth Surfaces. The only effect 
produced by the mutual pressure of two perfectly smooth 
surfaces would be the reaction perpendicular to them at 
the point of contact. It would hence exert no influence 
on the motion of one upon the other; in such a case 
there would be no friction. (There would still be, how- 
ever, even in this case, resistance to motion due to the 
mutual adhesion of the surfaces in contact; but this is 
an independent matter not here considered. ) 

An ideally smooth surface cannot be obtained, even 
by continued polishing; hence the resistance of friction 
can never be entirely eliminated. It is found in general 
that the smoother the surface is made the less is the 
friction. 

It is obvious, also, that the actual reaction between 
two rough surfaces which are in motion, or about to 
move, is in the direction of the resultant of the two 
components, one the force of friction parallel to the sur- 
face, and the other the normal pressure as defined in 
Art. 89. 



85.] FRICTION DEFINED. 9L 

84. Since friction is diminished by rendering the sur- 
faces in contact more smooth, it is customary to make 
use of lubricators, which fill up the unevennesses of the 
surfaces. For example, oils, lard, graphite, soapstone, 
and other substances are employed. The best lubri- 
cator, in a given case, depends upon the materials in 
contact and the amount of pressure sustained, and is 
determined by experiment. The friction of an axle 
may be much diminished by the use of friction-wheels; 
the axle rests upon two wheels, which turn with it, as 
indicated in Fig. 29 (p. 76) and Fig. 44 (p. 117). 

Conversely, when it is desirable to increase friction 
the surfaces may be made more rough. For example, 
when the driving-wheels of a locomotive tend to slide 
on the rails because the latter are wet and slippery, it is 
customary to feed down sand, by a tube from the sand- 
box above, on to the rail in front of each wheel, which 
has the desired effect. 

85. Friction is, so far as this, a disadvantage from a 
mechanical point of view, since force is required to over- 
come it, and this represents so much working power or 
energy expended without any useful result. For ex- 
ample, in the machinery of a cotton-mill, at every bear- 
ing there is friction, and the engine which supplies the 
energy for the establishment must furnish beyond what 
is required for the useful work performed enough more 
to make good this waste. Again, the locomotive on a 
railroad, after the train is in motion and supposing the 
track horizontal, exerts its energy solely against the out- 
side resistance, which is chiefly caused by the friction of 
the axles in their bearings and the wheels on the rails. 

On the other hand, however, friction is often mechani- 



92 DYNAMICS. [86. 

cally an advantage. It alone gives to the driving-wheels 
of the locomotive their hold on the track; without it 
the belts in a machine-shop could not be used to transmit 
the motion of one shaft to another; many mechanical 
arrangements depend for their efficiency upon it. Even 
so simple a matter as walking would be impossible but 
for the hold on the ground given to the feet by friction. 

86. Kinds of Friction. An important distinction is 
to be made between sliding and rolling friction. The 
former exists where there is simply sliding motion, as in 
the case of a sled, or of an axle in its bearing. The 
latter exists where motion is accomplished through the 
intervention of a wheel or roller. 

Sliding friction is that which is especially investigated 
here. The resistance due to it is much greater than 
that caused by rolling friction. This is seen by the 
effect produced when a carriage-wheel, by a shoe or 
some other contrivance, is made to slide instead of roll 
on the ground. The advantage of a wheel consists, as 
respects friction, in this, that instead of sliding friction 
on the ground, the rolling friction there and the sliding 
friction on the axle are substituted; the latter element 
is of comparatively small moment. Similarly, when 
heavy weights are to be moved for small distances, 
rollers of iron or wood are often placed under them. 

87. Fluid Friction. Friction, or resistance to motion, 
is also felt in the case of a liquid or gas, and is then 
called fluid friction. The resistance exists both between 
the molecules of the fluid itself and between them and 
the walls of the containing vessel. This is true, for 
example, wherfc water passes through pipes. This por- 
tion of the subject does not fall within the scope of this 



89.] 



LAWS OF FRICTION. 



93 



work, but the fact must be noted with reference to a 
subsequent article (104). 

88. Laws of Friction. Experiments upon friction 
haye established the following laws, which hold good for 
any two given surfaces in contact. It is supposed that 
no abrasion of these surfaces takes place. 

(1) The force of friction is proportional to the normal 
pressure of the surfaces in contact. 

(2) Friction is independent of the extent of surface 
in contact when the normal pressure remains the same. 

(3) Friction is independent of the Telocity of the 
motion when sliding friction is considered. 

89. (a) The normal, i.e. perpendicular, pressure (B), 
mentioned in the first law, is equal to the weight of the 




7r 



Fig. 33. Fig. 34. 

body if it rests upon a horizontal plane. Hence (Fig. 33) 

E = W. 

(i) If the body rests upon an inclined plane, the 
normal pressure is equal to that portion or component of 
the weight which is perpendicular to the surface. That 
is (Fig. 34), 

R z=W cos a. 

For if (Fig. 34) ac represents the weight of the body, 
then ad and ab are the two partial forces, or cornpo* 



94 



DYNAMICS. 



[90. 



nents, respectively parallel and perpendicular to the 
plane, into which it may be resolved (analogous to the 
resolution of velocities, Art. 38, p. 28, or see Art. 138, 
p. 144). Of these components, since bac = JJLK= a, 
. *. ad = W sin a, and this represents the force tending 
to make the body slide down the plane. Also, a b, i. e. 
IF cos a, is the pressure, normal to the plane, or the 
reaction of the plane as stated above. 

(c) If a force P acts on the body, at an angle ft, tend- 
ing to make it slide along, then the normal pressure is 
increased (or diminished) by that component of the 
force acting at right angles to the plane. If (Fig. 35) 



** 



iiictv L. rfi^IaHL 



r 

Fig. 35. 



vr 

t ■ 

Fig. 36. 



CA, exerted as a push, then DA = P sin ft, and 
R = W + P sin fi. 
A C, exerted as a pull, then AD = 



If (Fig. 36) P 
P sin /?, and 



R = W, - P sin ft. 



In either case, in order that the body shall be just at 
the point of moving, BA (Fig. 35) or AB (Fig. 36), i.e. 
P cos ft, must be just equal to the opposing force of 
friction. 

90. Explanation of the Laws of Friction. (1) The 

fikst law simply states that as this normal pressure 
(defined in a, b, c of the preceding article) is increased oi 



91.] 



COEFFICIENT OF FKICTION. 



AP 



Fia 37. 



diminished, the resistance of friction increases or dimin- 
ishes in the same ratio. This law can be demonstrated 
experimentally by an arrangement like that in Fig. 37. 
The weight W rests on a horizon- p w 
tal plane (so that R — W as in a). — 
A thread fastened to W passes 
over a pulley a, and is attached at 
the other end to a second weight 
P. In order that W shall be on 
the point of moving, P must be just equal to the force 
of friction (F). If W be doubled it will be found that, 
to satisfy the above condition, P must also be doubled, 
and so on. In other words, the ratio of P (= F) to 
W (= R) will remain constant; that is, the force of fric- 
tion is proportional to the normal pressure. 

(2) The second law may be illustrated by the case 
of a brick : supposing all the surfaces are alike in rough- 
ness, the friction is found to be the same upon which- 
ever of the three surfaces it rests. 

(3) The thikd law is generally but not rigidly true. 
It is found (1) that in the case of sliding friction the 
resistance to motion is a little greater when the body is 
just about to move, and (2) that in the case of very high 
velocities the friction becomes sensibly diminished. 

91. Coefficient of Friction. The constant ratio between 
the force of friction for two given surfaces and the normal 
pressure is called their coefficient of friction. If the 
coefficient of friction be represented by /x } then 

F 
V^-R* 



and 



F= }jlB. 



96 



DYNAMICS. 



[92 



M = -w> and F= /*W; 



This relation becomes on a horizontal plane 

F_ 

W 

on an inclined plane (89, h), 
F 



M 



and F = fi W cos a. 



Wcos a 

The coefficient of friction is a constant relation in a 
given case, depending only upon the nature of the sur- 
faces in contact. 

This use of the term coefficient, to denote a constant factor 
whose value depends upon the substance involved, is a very com- 
mon one in physical science. We speak, thus, of the coefficient 
of elasticity of a certain kind of steel; the coefficient of expan- 
sion, etc. 

92. Angle of Friction. The tangent of the angle of 
friction is equal to the coefficient of friction, /* = tan a. 




Fio. 38. 



I*et W be the weight of a body (Fig. 38) resting on a 
plane HL, and suppose that the plane is inclined at such 
an angle (a) that the body is on the point of sliding. 



93.] COEFFICIENT OF FRICTION. 97 

This angle is called the angle of friction, or angle of 
repose. 

The force of friction (F), acting in the direction aF, 
must be equal to the component of the weight, viz. ad 
(= TTsin a, as explained in Art. 89, b), which urges the 
body down the plane; that is, 

F = W sin a. (1) 

Also, the normal pressure, or the reaction of the plane 
aR, is equal to ab, the portion of the weight acting per- 
pendicular to LH\ but ab—W cos a (89, I). Hence 



Therefore 



and since u = 



R = W cos a. (2) 

F W Bin a 



R >Fcos a 



tan a; 



F_ 
R> 



pL = tan a. 



03. Determination of the Coefficient of Friction. The 

preceding article gives an accurate and simple method 
of obtaining the value of the coefficient of friction by 
experiment. If the surface of the plane is made of one 
of the materials in question, and that of the movable 
body in contact with it of the other, it is only necessary 
to observe (Fig. 38) the angle (a — HLN) at which the 
plane must be elevated in order to put the body on the 
point of sliding; then (92) tan a = u. 

The method illustrated by Fig. 37 (p. 95) may also be 
made use of. For, supposing the friction of the pulley 
to be so small that it can be neglected, if the weight W, 
representing one of the surfaces, is known, and also that 



98 DYNAMICS. [94. 

of P sufficient to put W on the point of moving on the 
other required surface, then the constant value of the 
ratio of these quantities is equal to the required co- 

efficient [~ = ~ = /*). 

94. Examples of the Coefficient of Friction. The 

limiting values of the angle of friction for different 
groups of substances, as determined by experiment in the 
case of sliding friction, are illustrated by Fig. 39. 



jb 




<^> V* 

Fig. 39. 

The corresponding range in the values of the coeffi- 
cient of friction is, as follows 

It, a (Angle of Friction.) 

Bricks, stones 0.60 — 0.73 31° — 36° 

Wood on wood 0.19 - 0.47 11° - 25° 

Metal on metal 0.14-0.22 8° - 12|° 

Lubricants 0.05 - 0.11 3° - 6° 

Many different circumstances affect these values ob- 
tained by experiment, so that the above are only to be 
taken as average results. 

For rolling friction the angle is much smaller. For example, it 
is stated that a railroad train in good order on a good road is not 



94.] FKICTION. 99 

safe against starting under the* action of gravity unless the gradi- 
ent is less than 18 to 20 feet to the mile (= 0° 13'); and that, if 
once started, the train will continue in motion on gradients as low 
as 13 feet per mile (Thurston). 



EXAMPLES. 
XVI. Friction. Articles 82-94. 

1. A force of 12 lbs. is just sufficient to move a body weighing 
48 lbs. uniformly along a horizontal plane: What is the coeffi- 
cient of friction? 

2. The value of n is .3, the weight of the body is 16 lbs. : What 
force is required to move it uniformly? 

3. It is found that a force of 7 lbs. suffices to move a body 
uniformly on a horizontal surface, where the value of the co- 
efficient of friction is known to be .25: What is the weight of the 
body? 

4. A body weighing 15 lbs. is just on the point of sliding when 
the surface it rests upon is inclined 20°: (a) What is the co- 
efficient of friction and the force of friction? (b) If the weight of 
the body is doubled, what values have these quantities? 

5. A body weighing 12 lbs. rests on an inclined plane whose 
angle of inclination is 14° and where// = .4: What is the force 
of friction? 

6. The ratio of the dimensions of an inclined plane are as 13 
(length) to 5 (height) to 12 (base): Will a body slide if the co- 
efficient of friction is (a) .4 and (l>) .5? (c) Yfhat is the force of 
friction in each case, the weight being 26 lbs. ? 

7. The dimensions of the plane are as in example 6 : What must 
be the value of the coefficient of friction if the force of friction on 
the plane is one half the weight of the body? 

8. A body weighing 12 lbs. rests on a plane where the co- 
efficient of friction is . 5 : What is the force of friction (a) if the 
plane is horizontal ? (b) if inclined 20° to the horizon? 

9. The length, height, and base of a plane are 30, 18, and 24 
feet : (a) What force is required to keep a body weighing 20 lbs. 
from sliding down, if ju = .2 ? (b) What force is needed to draw 
it uniformly up the plane? 



100 DYNAMICS. [94. 

10. A body weighing 50 lbs. rests on a horizontal plane, where 
jli = .2; What force is required to move the body uniformly if it 
acts as a, pull at an angle of 30° with the plane (Fig. 36)? 

11. What is the force required to move the body in example 10 
if it acts at the same angle but as a push (Fig. 35) ? 

12. A force of 60 lbs. acting as a pull at an angle of 20° moves 
a body uniformly on a horizontal plane, where ju = .3: What is 
the weight of the body? 

13. If the conditions in example 12 are fulfilled when the force 
acts at the same angle as a push, what is the weight of the body? 

14. A body weighing 4 lbs. is held against a rough vertical wall 
{fx = .6) by a force acting at right angles to the wall: What is the 
force? 

15. A force of 120 lbs. is just sufficient to support a body 
against a rough vertical wall (ju = .1); the force acts at right 
angles to the wall: What is the weight? 

[In the above examples no distinction is made between the re- 
sistance of friction when the body is just on the point of moving, 
and that which exists when the body is already in uniform motion ; 
in fact the former is sensibly greater than the latter.] 



CHAPTEE V.— WORK AND ENERGY. 

A. MECHANICAL WORK — MEASUREMENT OF WORK. 

95. Definition of Work. Work, in the sense in which 
the word is employed in Mechanics, is said to be done 
when a force acts upon a body and motion results in 
the direction of its action. 

There are two, essential elements here: (1) the force 
acting to overcome a resistance, and (2) the motion pro- 
duced. Where no motion results from the action of a 
force, no work is done; for example, a column support- 
ing a weight does no work. 

96. Examples of Work. Work is done against gravity 
when a man lifts a weight up from the ground, or when 
he ascends, i.e. lifts himself up, a hill; against friction 
when a horse draws a carriage along a horizontal road; 
against the molecular force of elasticity when a spring is 
wound up or a bow is stretched. 

97. Measurement of Work. The unit of work is 
the work done by a unit force acting through the unit 
of distance one foot. If the force be expressed in 
gravitation measure (72), then the unit of work is the 
foot-pound. 

(1) The work done by a constant force P, acting 
through a distance s, is equal to the product of the 
force into the distance; that is, 

the work done by P = P.s. (1) 



102 DYNAMICS. [97. 

If the force acts obliquely, at an angle /? with the 
direction of motion (Fig. 40), then only the effective 
component of the force, P cos /?, does the work. The 
work done is, therefore, 

P COS fi.B. (2) 

(2) The work done may also be measured by the 
^ effect produced; that is, when a 

weight W, expressed in pounds, 
is raised through a height h, ex- 




***-'" tS pressed in feet, the work done is 



Fig. 40. equal to the product of the weight 

raised into the vertical height ; that is, 

the work done in raising a weight = W.h. (3) 

The effective distance only — that is, the vertical height — 
is considered. If one pound is raised vertically one 
foot, then one foot-pound of work is done, and this is 
the simplest form of the unit of work as above defined. 

If the weight is an extended body, or if a number of bodies 
are considered together, the height to be taken is the vertical dis- 
tance through which the centre of gravity (denned in Art. 159) is 
raised. 

(3) Still, again, if a uniform resistance (R) is over- 
come through an effective distance d, then the work 
done is equal to the product of the resistance into the 
distance; that is, 

the work done against a resistance = R.d. (4) 

It is, in fact, immaterial, in the estimation of the 
amount of work done in any case, whether the attention 
be directed to the force acting, on the one hand, or the 
weight raised, or resistance overcome, on the other. For 



99.] MEASUREMENT OF WOEK. 103 

in all cases it follows from the law of the Conservation 
of Energy, as explained later (101 et seq.), that the two 
must be equal; that is, 

P.s = W.h, (5) 

or 

P.s = R.d. 

Whenever a weight is raised, the simplest method of 
estimating the amount of work done is by the product 
W.h. In some cases, however, it is more simple to 
measure the force acting and the distance through 
which it acts, and to obtain the number of foot-pounds 
of work done from the product P.s. 

98. When the distance through which P acts and W is 
raised differ, as when, for exam pie, by means of a set of pul- 
leys a small power acting through a great distance raises a 
large weight through a small height, the equation above 

p 

(5) shows that the ratio of -=? is the inverse ratio of the 

W 

P h 

distances through which they act, or — = — ; that is — 

The Power is to the Weight as the height through which 
the Weight is raised is to the distance through which the 
Power acts. This principle will be employed in the 
discussion of the various mechanical contrivances or 
machines, the lever, wheel and axle, etc., in Chapter 
VIII. 

99. Rate of Work. The rate of work is measured by 
the amount of work done, for example by a steam- 
engine, in a unit of time. The ordinary unit employed 
is called the horse-power, and is equal to 550 foot- 
pounds per second, or 33,000 foot-pounds per minute. 
This unit is employed in measuring the efficiency 
of a steam-engine. 



104 



DYNAMICS. 



(100. 



100. Applications of the above Principles. Suppose a 
constant force (P) acts to draw a body along a rough hori- 
zontal plane for a distance s. The work done is then 
equal toPxs [97 (1)]. But as it is difficult to determine 
the value of P directly, the principle may be made use of 
that: If the body moves uniformly, this force must be 
just equal to the force of friction; that is, P = F. But 
if the weight of the body is W and the coefficient of 
friction jj. is known, then F = }xR = }xW (91), and 

the work done against friction = jjiW.s. 

Suppose a weight W (Fig. 41) is raised to the top of 




Fig 41. 

an inclined plane whose height is h, and whose angle of 
inclination HLK is a; then (97) 

the work done in raising the weight = W.h. 

Also, if I is the length of the plane and fx the coefficient 
of friction for the surfaces in contact, since F= fj.R = 
jxWcosa (89, 91), 

the work done against friction = F.l= /*W cos a.l, 

and the total amount of work done is equal to 

Wh + yuTToos aX 



100.] WOEK. 105 

Since h = I sin a, this may be put in the form 

Wl (sin a + jucosa); 

. . b 

or again, since cos a — -r, 

Wh + pWb. 

The last expression shows that the work done in the 
case supposed is the same as that required to drag the 
body from L to K, and then to raise it yertically to H. 

EXAMPLES. 
XVII. Work. Articles 95-100. 

1. A weight of 600 lbs. is raised to tlie top of an inclined plane 
whose length is 1200 feet, and the angle of inclination = 10° : 
What work is done? 

2. A weight of 150 lbs. is raised to the top of a tower along a 
spiral path half a mile long, and which winds about it rising at a 
uniform angle of 15° : How much work is done on the weight? 

3. If a man in walking raises his centre of gravity a distance 
equal to ^ of the length of his step (as has been estimated), how 
much work will he do, if he weighs 150 lbs., in walking 20 miles? 

4. How much work is done by an engine weighing 20 tons in 
running a distance of a mile on a horizontal track, if the total re- 
sistance is 120 lbs. per ton? 

5. A sled weighing 1500 lbs. is dragged 10 miles on the snow, 
where the coefficient of friction is .075: What work is done 
against friction? 

6. How much work is done against friction in dragging a 
weight of 400 lbs. a distance of 1000 yards along a horizontal 
plane, if the coefficient of friction is .5? 

7. A weight of 250 lbs. is dragged up an inclined plane whose 
length is 2600 feet and the height is 1000 feet (ji = .3): How much 
work is done? 

8. A weight of 120 lbs. is drawn along a horizontal plane for a 



106 DYNAMICS. [101. 

distance of 1000 feet, and then up an inclined plane {a = 30°) for 
the same distance; the coefficient of friction is .2 for both sur- 
faces: What work is done? 



B. ENERGY — CONSERVATION AND CORRELATION OF 
ENERGY. 

101. Definition of Energy. Energy is the capacity of 
performing work. 

The first grand principle or doctrine of energy is called 
the Conservation oe Energy, which states that : 

The various forms of energy may be changed into one 
another, but the sum total remains the same; no energy 
is ever lost. This is a fundamental principle in all 
physical science, and its importance cannot be overesti- 
mated. It is, in a certain sense, a corollary from, and an 
extension of, the third law of motion (69). Like these 
laws its truth has been established by extended series of 
observations and physical experiments (66). 

As a proper understanding of this subject is neces- 
sary for a thorough comprehension of the laws of Me- 
chanics, and at the same time as this is impossible 
without a somewhat extended discussion of the subject, 
it is necessary at the outset to treat it broadly in its ap- 
plication to all Physics and not only as restricted to 
Mechanics. 

The term worh must, in the first place, be understood 
as having a wider signification than that given in the 
preceding portion of this chapter. Taken broadly, work 
is involved in the production of any physical or chemical 
change. For example, work is done not only when a 
mass of iron is raised from the ground, but also when by 
heat its temperature is raised; so, too, when water is 
changed from its liquid form into steam ; when an elec- 



103.] KINETIC AND POTENTIAL ENERGY. 107 

trical current is sent through a wire, as in the telegraph; 
and when the atoms of carbon of the coal unite with 
the oxygen of the air and combustion ensues accom- 
panied by heat and light. 

102. Forms of Energy. The forms of energy are 
divided into the following classes: 

(1) Mechanical energy, or the visible energy of 
masses of matter, including the energy due to elasticity. 

(2) Molecular energy, including heat, light, electricity, 
and magnetism. 

(3) Chemical energy, or that produced by the chemical 
union of unlike atoms. 

The forms of energy are also divided into (a) Kinetic * 
energy, or energy of motion, and (b) Potential energy, 
or energy of position. It will be sufficient here to apply 
this distinction more particularly to mechanical energy, 
but it also belongs to the forms which come under the 
other heads. 

103. Kinetic and Potential Energy. (1) Kinetic 
energy, mechanically considered, is that which belongs 
to bodies in virtue of their motion. Every moving body 
has a certain amount of energy, or capacity of perform- 
ing work, in consequence of this motion. For example, 
a swiftly moving cannon-ball, a running stream, the wind 
— all these, because of their motion, have a definite power 
of doing work. In fact, their motion is a result of energy 
expended upon them at some previous time, which they 
will themselves give back if their motion is arrested. 
This energy of motion is sometimes called vis viva, or 
" living force," and sometimes accumulated work. 

(2) Potential energy is that which belongs to bodies 

* From the Greek word hive go, to move. 



108 DYNAMICS. [104. 

in virtue of their position. Every body situated above 
the surface of the earth has a tendency to fall to it, 
under the action of gravity, and this determines its 
potential energy, or energy of position. For example, 
the weights of a clock, when wound up, have, because of 
their elevated position, a power of doing work ; e.g., in 
turning the machinery. So, too, a body of water 
behind a mill-dam represents a certain amount of poten- 
tial energy; that is, of energy which may be expended, 
e.g. in driving a mill if the water be allowed to fall. 
Also, the stones and bricks in a building represent a 
certain amount of energy expended once in raising them, 
and now present only potentially because of their posi- 
tion. A watch-spring when wound up, and a bow when 
stretched, are other examples of potential energy. 

In every such example the position of the body in 
question (or of its molecules), as the velocity in the pre- 
ceding case, indicates that a certain amount of energy 
has been expended upon it, and this, as before stated, 
will be given back on a return to the original position. 

The level of the sea may be made the surface of refer- 
ence for convenience; strictly, a terrestrial body would 
have potential energy anywhere except exactly at the 
centre of gravity of the earth. Moreover, other levels 
may be taken as the zero; for example, the weights of a 
clock, just mentioned, may be said to have no potential 
energy when they have fallen to the lowest point in 
their course. 

104. Measurement of Energy, (a) Inasmuch as the 
amount of work done — that is, in other words, energy 
expended — in raising a weight W through a vertical 
height h is equal to Wh, it is evident that: 



_^_ 



105.] KINETIC AND POTENTIAL ENEEGY. 109 

The potential eneegy of any body is equal to the 
product of its weight into the distance it has to fall. 

(b) Again : since, if a body of weight W fall through 
a height h, its energy of position is entirely changed 
into energy of motion, this last or kinetic energy will be 
equal to Wh. But in falling freely from rest through a 
distance h, a body acquires a velocity (27) such that 

v 7 = 2gh 9 and h = — . Substituting this value of h, the 

Wv* 
kinetic energy becomes — — . But the kinetic energy of 

any body of weight W and velocity v must be the same 
as that of the body which has acquired this velocity in 
falling. Hence, in general, the energy of a body in mo- 
tion is expressed by 

Wv* 

W 

Since the mass M = — (68), the value above may 

u 
be expressed in this form: 

M B 
-.,», or- 

The kinetic energy of any body is equal to the pro- 
duct of one half the mass into the square of the velocity. 

105. A body of weight TTand moving with a velocity 
v will, if brought to rest, expend against the resistance 

Wv % 

an amount of work equal to — — . If the resistance is 

uniform (as is true of friction), then, by the principles 
explained in Art. 97 (3), we shall have 

™L = JUL 

2<7 



110 DYNAMICS. [106, 

Wv 2 
Here — — is the amount of work accumulated, or 
2g 

stored up, in the moving body, and R. d is the work done 
against the resistance. If R is known, then d can be 
calculated, and vice versa. In the case of friction on a 
horizontal plane, for example, R = F '= fxW (91), and 
hence the distance a given body will slide can be easily 
computed. 

When d is very small, as when a heavy weight descend- 
ing drives a pile a short distance in the mud, then R, the 
average resistance, will be obviously very great. It is on 
this principle that, when a very great resistance has to be 
overcome, it is often most effectual to make use of the 
large amount of energy stored up in a moving body of 
considerable mass, which may be expended through a 
very small distance ; consider the efficiency of a heavy 
hammer. 

106. Relation of Kinetic Energy to Momentum. The 

distinction must be carefully made between the momen- 
tum and the kinetic energy of a moving body. Suppose 
the mass of the body to be M, and the velocity v ; then 

the momentum = M.v, 

the kinetic energy = -^-.v 2 . 

From these values it is seen that for the same body the 
momentum is proportional to the velocity, but the ki- 
netic energy — that is, the power of doing work — to the 
square of the velocity. For example, suppose two bodies 
to have the same mass, but let the velocity of one be 160 
feet per second, while that of the other is 80 feet per 
second; the former will have twice the momentum, but 



107.] KINETIC AND POTENTIAL ENERGY. Ill 

its kinetic energy will be four times greater. In other 
words, the first body can do four times as much work in 
coming to rest; it will ascend vertically upward against 
gravity four times as far; that is, 400 feet instead of 100 
feet. So, too, it will penetrate four times farther into 
a log of wood swung as that described in Art. 70 as a 
ballistic pendulum, though the momentum given to the 
moving mass will be only twice as great. 
In the relation in Art. 70, viz., 

MV ± mv — (M + m) v', 

it was shown that the momentum of the two bodies 
after impact was equal to the sum of their momenta 
taken separately before, they being supposed to be per- 
fectly inelastic. This is not true of the kinetic energy. 
Suppose the weights of two bodies to be 96 and 64 lbs. ; 
then M = 3 (=ff) and m = 2 (= ff): also, let V— 100 
feet per second, and v = 50; then, by the above equa- 
tion, v' = 80. Now the sum of the kinetic energies of 
M, (iMV*) and ira, (iwit; 9 ) is equal to 17,500 ft.lbs., but 
the kinetic energy of them moving together with the 
new velocity v' is only 16,000 ft.lbs. This difference 
would be still greater if the original motions were in 
opposite directions; viz., 17,500 ft.lbs. and 4000 ft.lbs, 
respectively. 

Hence there is an apparent loss of energy after impact. 
The equivalent of this loss is to be found in the heat 
produced by the blow, as explained further in articles 
108 et seq. This relation must be brought in in order 
to make Newton's third law of motion, in its general 
application, rigidly true. 

107. Transformation of Kinetic and Potential Energy. 

The two kinds of energy, considered in articles 103 and 



112 



DYNAMICS. 



[107. 



104, may be transformed the one into the other, and, if 
no other form of energy appears, the law of the Con- 
servation of Energy requires that this interchange 
should go on without loss. For example, suppose a 
ball (Fig. 42) is rigidly attached by a rod to the sup- 
port (7, about which it turns without friction. If its 

c 




Fig. 42. 

position be changed from B to A, and it be supported 
here for a moment, it is evident that a certain amount 
of work has been performed, or energy expended, upon 
it represented exactly by W.h, where li is the vertical 
height DB. This amount of energy has been imparted 
to the body, and belongs to it as potential energy in vir- 
tue of its position. 

ISTow suppose the ball to descend; at each successive 
point as M, in its course from A back to B, it loses part 
of its potential energy, but gains a corresponding amount 
of kinetic energy, or energy of motion; when B is reached, 

/ Wv 2 \ 

all its energy is that of motion I Wh = — — ). In virtue 



of this motion it will ascend on the other side, exchang- 
ing at each point its energy of motion for energy of 
position, and at A' its velocity is and its energy all 
potential. 



107.] KINETIC AND POTENTIAL ENEKGY. 113 

Again, on descending, the exchange of energy takes 
place as before. If the supposed conditions of a per- 
fectly rigid rod, moving without friction at C and meet- 
ing with no resistance of the air, could be realized, this 
motion would go on forever; it would be one kind of 
perpetual motion. [It would not be the "perpetual 
motion" sought for ; for example, an engine which shall 
go on doing work forever without being supplied with 
fuel; this the doctrine of energy shows to be absurd and 
impossible. ] 

Another example will further illustrate the transformation of 
the two forms of energy. Suppose a ball to be projected from 
the ground vertically upward, with a velocity v ; the energy at 
the moment of starting is all kinetic. By the laws of kinematics 
(44) it will, if gravity alone resists it, ascend to a height (A) so 

that h = — -. As it ascends its kinetic energy is continually ex- 

changed for energy of position, and at its highest point it has only 
potential energy. If its weight is W, the amount of work done is 

Wh, but h = — -; hence the work done, which is the equivalent 

%g 

of the initial kinetic energy, is -g— (same result reached as in Art. 

"9 
104, b). Further, on its descent it will continually exchange its 
energy of position for that of motion, and when it reaches the 
ground its energy is all kinetic. Suppose that the ball and the 
surface it comes in contact with are perfectly elastic. The result 
of the blow will be to compress the molecules of the body for an 
instant, and at this moment it is at rest and its energy is repre- 
sented potentially by the new position of the molecules. In con- 
sequence of the elasticity, however, the molecules tend to regain 
their original position, and thus the potential energy is again 
transformed into kinetic, and the effect is to project the ball up- 
ward with the same velocity as before. In the ideal case supposed 
this exchange would go on continually, and the ball would bound 
forever. 



114 DYNAMICS. [108, 

108. Apparent Loss of Visible Energy. It is obvious 
that the conditions supposed in the preceding article 
cannot be realized, but that the pendulum and the 
bounding ball will sooner or later come to rest. In such 
cases there is an apparent loss of visible energy. Still 
more is there an apparent loss of energy when the 
motion of a train is arrested at a station, or that of a 
cannon-ball by a target. In such cases it was once be- 
lieved that the energy was really lost, but it is now 
known that this is as untrue as it would be to suppose 
that the matter in a piece of paper is lost when it is 
burned. The matter is unchanged in amount, and is 
truly indestructible, though its form may be altered and 
it so become invisible to the eye. In an analogous way, 
in this apparent loss of visible energy, a new form of 
energy takes the place of that which disappears, for the 
energy itself is indestructible. This form of energy, 
produced as the equivalent of the mechanical energy 
which has disappeared, is heat, 

109. Nature of Heat. Heat is now believed to be, 
not a form of matter, as once supposed, but a "mode 
of motion ;" more particularly it is a very rapid un- 
dulatory vibration of the particles of matter making up 
the heated body. When heat is transmitted through a 
medium without raising its temperature it is said to be 
radiated, and the undulatory motion is believed to be 
propagated at a very great velocity by the particles of a 
supposed elastic fluid called the ether. Thus, the heat 
of a stove is said to be radiated in all directions from it; 
so, too, the heat of the sun is said to be radiated to the 
earth, and the heat received is called radiant Jieat, or 
radiant energy. 



111.] MECHANICAL ENERGY AND HEAT. 115 

When, however, the heat is transmitted through a 
body at a comparatively slow rate, as from one end of 
an iron rod thrust in a furnace to the other, it is said to 
be conducted, and in this case the particles of the bar 
itself are believed to propagate the motion. 

A hot body is one whose particles are in rapid motion; 
but "hot," as the word is used, is only a relative term, 
for this motion belongs to the molecules of all bodies of 
which we have any knowledge, however "cold," and the 
rapidity of the motion determines the degree of heat 
(temperature) as manifested, for example, to our senses 
or to a thermometer. 

110. Examples of the Production of Heat from Me- 
chanical Energy. Examples of the appearance of heat 
at the same time with the disappearance of visible 
energy are very common. Thus, a nail rubbed quickly 
with a file becomes " hot;" the same is true of a metallic 
button rubbed on a piece of cloth. A piece of iron on 
an anvil may be raised to a dull red heat by rapid blows 
from a hammer; a friction-match is ignited by the heat 
produced by a scratch; a cannon-ball whose motion is 
arrested by a target is itself, as well as the target, very 
much heated by the collision. In all such cases, as will 
be apparent from what has been said, the visible mass 
energy is exchanged for the molecular energy of heat: 
the slow motion of the mass for the very rapid motion 
of the molecules. 

111. Definite Relation between Heat and Mechanical 
Energy. If it be true that the heat produced in the cases 
named (110) is the equivalent of the visible energy lost, 
it follows that there must be a definite numerical ratio 
between a certain amount of work done and the amount 



o* 



116 DYNAMICS. [111. 

of heat produced by it. This relation has been estab- 
lished in many different ways by different experimenters, 
but in all the cases the essential part of the process is 
the same, viz., to measure the amount of mechanical 
energy expended (in foot-pounds), and also to determine 
the amount of heat produced as its equivalent. 

Heat is measured in heat-units; that is, the unit of heat is 
y £ that amount of heat required to raise one pound of 
water one degree in temperature. For physical prob- 
lems the Centigrade thermometer is universally em- 
ployed; but with English-speaking people the Fahren- 
heit thermometer is commonly used as the house 
thermometer. The relation of the two is evident from 
Fig. 43. For the Centigrade thermometer the freezing- 
point of water is made the zero, and the distance from 
it to the boiling-point is divided into 100 degrees. In 
the Fahrenheit thermometer the freezing-point is 32 de- 
grees above the zero, and the boiling-point 212 degrees 
above. Hence 100 degrees Centigrade correspond to 
180° (= 212° - 32°) Fahrenheit. To change the read- 
Fig. 43. i n g S f either thermometer to those of the other, we 

have representing the number of degrees in the two scales by F 

and G respectively, 

£<7+ 32° = F, and %(F- 32°) - G 

The method which was employed by Joule* was es- 
sentially as follows : A metal box B (Fig. 44) was taken 
full of water; in this was placed a paddle (Fig. 45) at- 
tached to a vertical spindle A, which could be revolved 
by means of a string passing over two pulleys C and D, 
and attached to two known weights E and F. If now 

* The experiments of Dr. Joule were carried on between 1843 
and 1849. He employed several different methods for determin- 
ing the " mechanical equivalent of heat," but that which led to 
the most satisfactory results is the one here described. 



1 



111.] 



MECHANICAL ENERGY AND HEAT. 



117 



these weights (E + F — W) are allowed to descend 
freely through a distance h, marked on the vertical 
scales, the work done is Wh; but this is expended in 
turning the paddle in the water, and, owing to the fric- 
tion of the water (Art. 87), is all transformed into heat. 
Hence if the amount of water is known, and its tempera- 
ture before and after the experiment, the number of 
foot-pounds of work required to produce one heat-unit — 
that is, to raise 1 lb. of water 1° C. — can be readily 
calculated. In the actual experiment it was necessary 
to make corrections for the loss of energy in the friction 




Fig. 44. 



Fig. 45. 



x>f the pulleys, the radiation of heat from the box, aud 
several other points which need not be explained here. 
The result obtained was this: That an expenditure of 
1390 foot-pounds of work produce one unit of heat on the 
Centigrade scale (772 ft. lbs. on the Fahrenheit scale). 
Many other experiments have been made in various ways 
having as their object the determination of this same 
relation; for example, the amount of heat produced by 
the friction of two iron plates in mercury, that caused 
by the collision of two heavy bodies one of which has 



118 DYNAMICS. [112. 

fallen through a known height, and so on. All these 
experiments have confirmed the relation obtained by 
Joule. 

112. Conversion of Heat into Work. Since a definite 
amount of mechanical work is equivalent to a certain 
amount of heat energy, the converse must also be true: 
that heat is convertible into mechanical work. This 
conversion of the former kind of energy into the other 
is best seen in the steam-engine. Here the burning coal 
in the furnace converts the water of the boiler into 
steam, and the expansion of this steam in the steam- 
chest connected with the condenser sets the piston in 
motion backwards and forwards. This is mechanical 
motion, and it may be utilized, for example, to drive 
the lathes in a machine-shop, to pump up water, to 
propel a steamship or a train of cars. All these are 
cases in which from heat mechanical work is obtained. 

There is one most important difference between the 
two cases that have been described; viz., the transforma- 
tion of work into heat, and that of heat into work. The 
former transformation can be completely made, but 
under no conditions, which are practicably obtainable, 
can a certain amount of heat be all changed into me- 
chanical work. A perfect "reversible engine" (as that 
of Carnot) can be conceived of, but the necessary con- 
ditions cannot in practice be even approximately realized. 

113. Other Forms of Molecular Energy. The form 
of molecular energy most closely related to Mechanics 
is that just considered; viz., heat. There are, how- 
ever, other forms of energy into which mechanical work 
may be converted, and conversely from which work may 
be obtained. Here belong the other physical agents, 



114.] TKANSEOKMATION OF ENEKGY. 119 

light, electricity, magnetism. The full understanding 
of their mutual relations requires an extended knowl- 
edge of Physics, and cannot be attempted here, but some 
illustrations will suffice to make the subject clear. 

114. Examples of Transformation of Energy. The 

production of Avork from burning coal, already given, is 
one example. Another example is this: The water in 
a mill-pond represents a certain amount of potential 
energy (103); this in falling through the mill-race may 
be made to turn a mill-wheel, and its motion represents 
a certain amount of mechanical energy derived from the 
water; if a turbine wheel is employed, from 60 to 80 per 
cent of the energy of the water may be utilized under 
favorable conditions. The motion of the wheel may be 
given to a saw, which shall do work in overcoming the 
cohesion of the wood, or to millstones, by which grain 
is ground. Or, again, it may turn an electro-magnetic 
machine. This cannot be described here, but it is suffi- 
cient to understand that the machine accomplishes this : 
that the rapid mechanical motion results in the produc- 
tion of a current of electricity, which is the equivalent 
of a certain portion of the mechanical energy. This 
electrical energy may be conveyed by a copper wire for a 
considerable distance (with a loss, however, for some of 
it is inevitably transformed into useless heat), and then 
the electricity may be used to make a light, when the 
remainder of the energy is transformed into light and 
heat, or it may be used to do work in chemical separa- 
tion, as in electro-plating with copper. Still, again, it 
may by means of a second similar machine be trans- 
formed back again into mechanical motion, and this 
used to do any kind of work desired. 

Again, the revolution of a disc of plate-glass between 



120 DYNAMICS. [115. 

cushions suitably arranged may be made to produce 
electricity, which, is the equivalent of part of the me- 
chanical energy expended in turning the wheel, the 
remainder being expended against friction and resulting 
in heat. If this electricity is collected on a brass cylin- 
der and then a spark taken from it, the light and heat of 
the spark, with the vibratory motion resulting in the 
noise, are the forms of energy into which the mechanical 
work has been transformed. 

Chemical affinity also represents a most important 
kind of energy. Two dissimilar atoms under suitable 
conditions combine, and some other form of energy is 
the result. Thus, in the combustion of coal in air, it is 
the union of the carbon and hydrogen of the coal with 
the oxygen of the air which results in the formation of 
heat energy. So, too, the charcoal, nitre, and sulphur 
mixed together in gunpowder will unite under proper 
conditions, and the result is the appearance of energy 
which produces not only heat and light (and noise), but 
also may do a vast amount of mechanical work. 

115. Conservation of Energy. To all the examples of 
the transformation of one form of energy into another, 
given in the preceding article, the law of the Conserva- 
tion of Energy applies. It requires that the sum total 
should remain the same, that no energy should be lost 
in the changes. In order to prove this rigidly it would 
be necessary to be able to correlate all the different 
kinds of energy and express between them a definite 
ratio, as that between heat and mechanical energy. This 
cannot always be done, for of the real nature of some of 
these forms of energy but little is certainly known; but 
physical investigations have gone so far as to make it 
sure that the fundamental principle here stated is true. 



117.] ENERGY DERIVED FROM THE SUN. 121 

116. Terrestrial Stores of Energy. From the dis- 
cussion in the preceding articles it appears that, for the 
performance of the many kinds of work necessary for 
human life on the earth, the best possible use must be 
made of the various forms of energy which are available, 
for these cannot in any way be increased. 

The most important stores of energy, from which 
mechanical work can be obtained, are the following: 

1. Energy of water either potential or kinetic: this is 
utilized by means of the various water-wheels, and made 
to drive mills, etc. This includes the energy of tidal 
water, which is also occasionally made use of. 

2. Energy of wind: employed to do work in propelling 
ships, and in turning windmills. 

3. Energy of coal, wood, oil, and other combustibles: 
utilized as fuel principally in the steam-engine. 

4. Energy of the muscular effort of the various 
animals, including man. 

5. Direct energy of solar heat and light radiation : it 
has been found possible to employ this in running a 
solar engine, but thus far no extensive use has been 
made of it. The indirect way in which solar heat and 
light have been and are still being utilized is mentioned 
in the next article. 

To the above may be added the energy of uncombined 
chemical elements, as sulphur and iron; also, the internal 
heat of the earth; the earth's rotation (note the remark 
at the close of the next article on tidal energy); finally, 
the potential energy of masses of matter above the mean 
surface of the earth, which have been elevated by 
geological changes in the past. 

117. The Sun as the Ultimate Source of Terrestrial 
Energy. All of the important forms of energy just enu- 



122 DYNAMICS. [117. 

merated (116) are derived either directly or indirectly 
from the sun. The sun is constantly radiating out in 
all directions into space a vast amount of heat and light 
energy. Of the whole amount but a very minute frac- 
tion is received by the earth, and only a very small part 
of this is utilized, but this relatively small amount is 
essential to the existence of all kinds of life on the 
earth. 

1. The heat energy of the sun causes evaporation from 
every sheet of water; the water thus raised in the form 
of vapor falls again on the earth's surface, as rain or 
snow, much of it at a level far above that of the sea. It 
forms running streams, or is collected in lakes and 
ponds. In descending again to the sea-level it may be 
made to do a great amount of work. 

2. The heat energy of the sun is, also, the chief cause 
in setting the air in motion in the form of winds, and 
these, as have been stated, drive our ships and turn our 
windmills. 

3. Still more important, the heat and light energy of 
the sun are the cause of all vegetable growth; that is, 
under their combined action the chemical change goes on 
by which the carbon (from the CO2 in the atmosphere) 
is built up into the structure of the plant or tree. The 
energy thus appropriated is stored up, but it may be ob- 
tained again, chiefly in the form of heat, when the wood 
is burned as fuel. The accumulated vegetation of a 
former and far-distant period has been changed, though 
without any considerable loss of energy, to coal, and 
this therefore now represents potentially the energy of 
the sun received and utilized by the earth at that 
time. When, now, the coal is burned in the fire-box of 
a steam-engine, this long-stored-up potential energy be- 



118.] DISSIPATION OF ENERGY. 12o 

comes again kinetic, and from it we obtain a large part 
of our mechanical work. 

Of the whole amount of the sun's energy which has 
its equivalent in the resulting vegetable growth, part, as 
has just been said, is obtained again in the combustion 
of fuel. Another part is obtained again indirectly 
through the muscular work of animals, who have used 
the vegetable growth in one form or another as food; for 
an animal, as regards its capacity for performing physi- 
cal work, is to be regarded as a machine for the trans- 
formation of energy, which must be fed with fuel as 
truly as the steam-engine. A man forms no exception 
to this statement, for, in order that he may live and do 
work, he must also be fed with fuel; in his case, how- 
ever, the process is one step more complex, since his 
principal food is the flesh of animals, who themselves 
have derived their support from vegetable growth. 

The energy of the tides must be made an exception to 
the preceding remarks, for they are due to the attraction 
of the sun and moon; the energy of the tides may, in 
fact, be shown to be derived in part from the energy of 
the earth's rotation, the rapidity of which they conse- 
quently tend to diminish to a very small extent. 

118. Dissipation of Energy. If the illustrations of 
the transformation of energy which have been given be 
carried out one step farther than is attempted, and if, 
too, the attempt is made to apply the principle of the 
Conservation of Energy to them rigidly, it will be seen 
that at each step in every transformation there is an appa- 
rent loss of energy (a real loss as regards useful energy) 
by its change into useless heat, and, moreover, that the 
final form which it tends to assume is always that of heat. 

For example, a pound of coal produces upon combus- 



124 DYNAMICS. [118. 

tion in the air, by the change of chemical energy into 
heat energy, about 7500 heat-units; but 1390 foot-pounds 
of work are the equivalent of 1 heat-unit; hence 1 pound 
of coal should afford 7500 X 1390 = 10,425,000 foot- 
pounds of work. But the best steam-engines utilize only 
about 10 per cent of this; the remainder is lost — not, in- 
deed, as energy, but so far as useful effect goes. Moreover, 
of this small fraction utilized, for example, in the case of 
the energy required to set a train in motion and to carry 
it to the next station, much of this is expended on the 
way against friction (i. e. , is transformed into heat), and 
when the brakes are applied and the motion of the great 
mass is arrested, the remainder of the energy is also all 
transformed into heat. The same is true for the other 
examples given. 

The heat, however, which is thus produced as the 
last step in these transformations is heat at a low tem- 
perature, which we are unable to utilize and which is 
virtually lost. For as work can be obtained from a body 
of water only as it descends from a higher level to a 
lower, and if all were at the lower level, no work would 
be possible; so work can be obtained from heat only as 
there is a passage from heat at a higher to that at a 
lower temperature. If all bodies had a uniform tempera- 
ture, no work would be possible. There is then a second 
law : that of the Degradation of Energy : according 
to which useful energy is being constantly exchanged for 
heat of which no use can be made. 

EXAMPLES. 
XVIII. Potential and Kinetic Energy. Articles 101-118. 

1. The weights of a clock weigh 60 lbs. and they have 30 feet 
to fall: How much work do they represent when wound up? 



118.} ENEKGY. 125 

2. A mill-pond has a surface of 1 acre and an average depth 
of 6 feet; suppose it 200 feet above the sea-level : How much 
potential energy does it represent (1 cubic foot water = 62.5 lbs.)? 

3. If the amount of water which passes over a waterfall 150 
feet high is 1000 cubic feet in a minute, and if the energy derived 
from the fall alone could be utilized, how much work-power 
would it represent? 

4. If the same amount of water, as in example 3, passes through 
rapids above the fall at an average velocity of 16 feet per second, 
how much additional kinetic energy is here present? 

5. Suppose that the energy of the water in 3 and 4 is all ex- 
pended by the impact at the bottom of the fall : How much heat 
will be generated, and how much would the temperature of the 
water be elevated? 

6. How much work is accumulated or stored up (= kinetic 
energy) in a cannon-ball weighing 200 lbs. and moving at a rate 
of 1200 feet per second? How much heat will be generated if 
its mass motion is entirely destroyed by the impact with the 
target? 

7. An ounce bullet has a velocity of 800 feet per second : How 
much work can it do ? 

8. (a) How high will the ball in example 7 ascend if moving 
vertically upward? (b) How much farther if its velocity is 
doubled? 

9. If an arrow will ascend a certain distance vertically, how far 
will a second of twice the weight ascend, the initial velocity being 
the same? 

10. A body weighing 20 lbs. is projected along a rough hori- 
zontal plane (jn = .25) with an initial velocity of 320 feet per 
second : How far (a) will it go before coming to rest ; how long 
(b) will it slide ; and at the end of 2 seconds how far (c) will it 
have gone, and (d) what will be its velocity? 

11. A body weighing 80 lbs. is projected along a rough hori- 
zontal plane with an initial velocity of 200 feet ; the coefficient of 
friction is £: (a) How far and (b) how long will the body con- 
tinue to move, and (c) how much work is done against friction in 
10 seconds? 

12. A body weighing 100 lbs. slides down an inclined plane 
whose length, height, and base are respectively 100, 60, 80 feet. 



126 DYNAMICS. [118. 

and with the velocity thus acquired ascends another inclined 
plane whose dimensions are 100, 80, 60 feet respectively; the coeffi- 
cient of friction is .1 . How far will it go, the change of direction 
being supposed to take place without loss of velocity? 

13. The body in example 10 is projected up an inclined plane 
(ju = ,25) whose length, height, and base have the ratio of 
10:6:8 (a) How far and (b) how long will it ascend? 

14. An inclined plane has the dimensions, length 2000 feet, 
height 1600, base 1200, and jn = .2: What velocity of projection 
must a body weighing 64 lbs. have just to reach the top? 

15. If a body starting from rest slides down the plane in example 
14, (a) how much work will be stored up in it when it reaches the 
bottom? Also, (b) how far will it slide on a horizontal surface 
(M = .2) if the change in direction occasions no loss in velocity? 

[In the following examples the kinetic energy is supposed to 
be expended against the resistance alone; in fact, a considerable 
portion would result in the immediate production of heat. More- 
over, the resistance is supposed to be uniform ; in fact, this is not 
the case, and the result obtained in each problem consequently is 
only the average resistance. ] 

16. A hammer weighing 12 lbs. and moving with a velocity of 
4 feet per second drives a nail into a plank half an inch : What 
resistance does it overcome? 

17. A weight of 1000 lbs., used as a pile-driver, falls 20 feet, 
and drives the pile in one inch: What resistance does it overcome? 

18. Two balls weighing 100 lbs. each are attached to the ends 
of a horizontal bar, this is attached to a screw of rapid pitch 
(236). They are made to rotate rapidly and have a velocity of 
10 feet per second, when the end of the screw strikes the metal 
to be stamped. Suppose that the punch comes to rest after 
moving through ^ inch: What resistance is overcome? 



w^^mtm^ 



CHAPTER VI.— STATICS. 
Introductory. 

119. Statics is that branch of Mechanics which con- 
siders the action of forces in so far as the body acted 
upon is held by them in equilibrium (61). 

120. Geometrical Representation of a Force. A force 
may be represented geometrically by a straight line in a 
manner analogous to the graphic representation of Telo- 
city (20). In each case (1) the point of application, (2) 
the direction, and (3) the magnitude of the force are 
supposed to be known. 

Thus, Fig. 46, (1) the position of the particle A acted 
upon represents the point of application of 
the force; (2) the direction of the line AB 
represents the direction of the force — that 
is, the direction in which it tends to move 
the particle A; (3) the length of the line £, 
AB represents the magnitude of the force, fig. 46. 
being taken proportional to this magnitude in terms of 
an adopted unit. It is obvious that the length of the 
line has meaning only when the unit of comparison is 
known, or when two or more forces are represented by 
lines in terms of the same unit; in the latter case the 
ratio of the lines in length will be also the ratio of the 
forces in magnitude. 

In statics a force is usually exerted as tension or pres- 
sure, and is measured in pounds; that is, in accordance 




128 STATICS. [121. 

with the gravitation method (72), by the weight it could 
support. The unit force is equal to the force of attrac- 
tion exerted by the earth upon one pound of matter. 
The possible error involved in this method has already 
been explained (72). 

121. Line of Action of a Force. The line of action of 
a force is the line in which it acts, taken irrespective of 
its direction. Thus, Fig. 46, the line of action of the force 
represented is the indefinite straight line AB (or BA) 
produced; the direction of the force is from A to B, 
which is always indicated by the order (AB) in which 
the letters are named, often also by an arrowhead. 
Forces may therefore have at the same time the same 
line of action but opposite directions. 

122. Transmission of a Force in its Line of Action. It 

may be accepted without demonstration that if a force 
act upon a body at any point, it may be applied without 
change at any other point in its line of action which is 
rigidly connected with it. Conversely, if a force may be 
transmitted without change from one point to another 
rigidly connected with it, the second point must lie in 
its line of action. 

It is also true that a force is transmitted by a string 
without loss, even if the direction of the string is 
changed by its passing over pegs or pulleys; in other 
words, the tension of a string is the same at every point. 
In this statement it is assumed that the string is per- 
fectly flexible, and that there is no friction. 

123. Body; Particle. The definitions of a preceding 
article (2) are here repeated: A body is a portion of mat- 
ter having definite dimensions. A particle is a portion 
of matter so small that any difference in the position of 



125.] COMPOSITION OF FOECES. 129 

its parts may be left out of account. All principles 
established for forces acting on a particle will be true 
also for forces acting at a point of a body, or for forces 
whose lines of action produced will pass through the 
same point in the body (122). A body is always sup- 
posed to be perfectly rigid. 

Composition of Forces meeting at a Point. 

124. Composition of Forces. Components and Ke- 
sultaxt. The single force which will produce the 
same effect as two or more forces acting together is 
called their Resultant. The individual forces them- 
selves, which may be thus combined, are called the Com- 
ponents. 

The process of finding the resultant of two or more 
forces in direction and magnitude is called the Composi- 
tion of Forces. 

125. General Condition of Equilibrium. A particle 
will be in equilibrium when the forces acting upon it 
balance one another so that their resultant is zero. The 
general condition of equilibrium, therefore, for any 
number of forces acting on a particle, or at a point of a 
body, is 

R = 0. 

In order that this shall be satisfied various special con- 
ditions must be fulfilled, in the different cases which 
arise, as mentioned hereafter. 

It is obvious that of several forces acting on a particle 
and holding it in equilibrium, each force will be equal 
and opposite to the resultant of all the others. For the 
remaining forces can be replaced by their resultant 
without change of effect, and in order that there shall 



130 STATICS. [126. 

be equilibrium the first force must be, as stated, equal 
and opposite to this resultant. A body in equilibrium 
must be acted upon by at least two forces ; a single force 
always causes accelerated motion (60). 

126. Composition of Forces having the same Line of 
Action, (a) The resultant of two forces, or of any num- 
ber of forces, acting in the same line and in the same 
direction is equal to their sum. For example, if P, Q, 
8, T, etc., represent forces haying the same direction, 
as, e.g., those exerted by several men pulling in the 
same line on a rope, and R the resultant, or equivalent 
single force, then 

B=P+Q+S+T+ etc. 

Again, (b) the resultant of several forces acting in the 
same line, but some in one and others in the opposite 
direction, is equal to the sum of the first subtracted 
from the sum of the second; and it will act in the direc- 
tion of the greater sum. If the respective directions of 
the forces be distinguished by their algebraic signs 
(+ or — ), then the resultant will be equal to the alge- 
braic sum of all the forces, and its direction will be 
indicated by its sign. 

127. Condition of Equilibrium for Forces having the 
same Line of Action. The condition of equilibrium for 
two or more forces having the same line of action is this: 
their algebraic sum must be equal to zero. 

128. Composition of two Forces not having the same 
Line of Action: Parallelogram of Forces. If two 
forces acting on a particle be represented in direction 
and magnitude by the two adjacent sides of a parallelo- 
gram, then the diagonal of this parallelogram passing 



128.] COMPOSITION OF FORCES. 131 

through their point of intersection will represent th6 
magnitude and direction of the resultant. 

This proposition is at once seen to be closely similar to 
the Parallelogram of Velocities (33 and 39). That prin- 
ciple, as has been stated (68, b), is a deduction from the 
second law of motion, and the same is true of the Paral- 
lelogram of Forces. The part of that law upon which 
both propositions are based may be stated in this form: 
When several forces act simultaneously upon a body, each 
force produces exactly the same effect which it would 
have produced if it had acted singly. This principle is 
true whether the body was originally at rest or in mo- 
tion, and it extends as well to the case where the forces 
balance one another, so that the body is in equilibrium. 

In applying the principle we may consider the forces 
either (a) dynamically, as producing motion, or (b) stati- 
cally, as causing pressure or tension without motion. 




In the first case (a), suppose two forces P and Q to act 
simultaneously upon the same particle; each will have 
the same effect as if it acted alone, and (71) it is meas- 
ured by the Telocity it gives in a certain time, and its 
direction is that of this Telocity; therefore if these velo- 
cities are represented (Fig. 47) by AB, AD respectively, 
then the same lines must be proportional to the forces 
P and Q; further, since AC represents the resultant 



132 



STATICS. 



[129. 



Telocity in direction and magnitude, a force haying this 
direction and proportional to A C must be the resultant 
force equivalent to the combined effects of P and Q. 
Hence the geometrical methods of compounding forces 
are the same as those of compounding velocities. 

In the second case (b), the forces are generally meas- 
ured by the weights they can support (72), but the same 
geometrical methods are also applicable to them. An 
experimental proof of this latter case is given in the next 
article. 

129. Experimental Verification of the Parallelogram 
of Forces. Let (Fig. 48) A and B be two pulleys, 



i ®<? 




Fi». 48. 



whose position may be changed at will; over these are 
stretched two silk threads, knotted at a, sliding without 
friction. At the extremities of these threads are hung 
two weights P and Q, and from a is hung a third 
weight, found by trial to be just sufficient to balance 
P and Q for the given position of A and B. The parti- 
cle a is now in equilibrium under the action of the 
three forces P, Q, and W, and it is obvious that the 
resultant of P and Q is equal and exactly opposite to 
IT (125). 



130.] 



COMPOSITION OF FOKCES. 



133 



If now, from a, ab be measured off, containing as many 
units of length as there are units of weight in Q, and 
also ad, containing as many units of length as there are 
units of weight in P, these forces will be represented in 
magnitude by ab and ad respectively, for the pulleys 
change their directions only. Complete the parallelogram 
abed; it will be found on trial that the diagonal ac, which 
by the proposition must represent the resultant of P and 
Q, is vertical — that is, directly opposed to W — and also 
that it contains as many units of length as there are 
units of weight in W; hence the proposition is true in 
this case. 

If the positions of A and B be changed, and also the 
magnitudes of P and Q, and a weight W be hung at a 
which will hold the system in equilibrium, it will be 
found in every case that the proposition is verified in 
the same manner as above, and hence it may be accepted 
as always true. 

130. Calculation of the Resultant. General Case. Let 
the two forces P and Q, whose directions form any 




angle y with each other, be represented (Fig. 49) by 
AB and AD respectively. Then, if the parallelogram 
ABCD be completed, the line A C will represent the re- 
sultant of these forces. It is required to find an expres- 



134 



STATICS. 



[131. 



sion for the magnitude of this resultant in terms of 
P, Q, and y. 

From geometry (Fig. 49), 

AC = AB* + BC 9 + 2AB.BE, (1) 

but BE = BC cos CBE = AD cos DAB. 

Therefore, substituting the values of AB, AD, BE in 
(1), we haye 

B* = P 2 + Q* + 2PQ . cos r . (2) 

The same formula may be shown to hold true for any 
other case, as when (Fig. 50) the angle y is obtuse. 

D C 

\ 

1 9 




Further (Figs. 49, 50), since BC — AD — Q, it is seen 
that the relations between the two forces and their 
resultant are equally well expressed by the triangle ABC. 
This triangle is often useful for calculation, for (Figs. 
49 and 51) AB = P, BC = Q, AC = R, BAC = a, 
ACB = DAC = fir, and ABC = (180° - DAB) = 
(180° — y). Therefore all the relations between the 
two forces and their resultant, in magnitude and direc- 
tion, may be calculated from the triangle ABC by the 
ordinary methods of solving an oblique-angled triangle. 

131. Special Cases, (a) P = Q. The general value 
of R in (2) above (130) becomes, if P = Q (Fig. 52), 



131]. 



COMPOSITION OF FORCES 




P 2 = P 2 + P 2 + 2P 2 cos y, 
= 2P 2 + 2P 2 cos y, 
= 2P 2 (1 + cos y). ** 



But, by trigonometry, cos \y =y — ^~ — -, or 
1 -f- cos y = 2 cos 2 \y\ therefore 

P 2 = 2P 2 (2 cos 2 \y\ 
= 4P 2 cos 2 1x5 
.-. P = 2Pcos|^. 

This result, as also those of (b) and (c) below, may be 
obtained directly from the figures without reference to 
the general case. It is seen here that the resultant of 
two equal forces bisects the angle between them. 

(b) p = Q f y =£= 60°. The general value of P becomes 
for this case 

P 2 = P 2 + P a + 2P 2 cos 60°, 
= 3P 2 ; 
.-. P = PVs. 

(c) P = Q, y = 120° (Fig. 53). The general value 
of P becomes in this case 

P 2 = P 2 + P 2 -j- 2P 2 cos 120°, 
= 2P 2 - P 2 ; 

s. R = P. 




136 



STATICS. 



[132, 



This result shows that, if two forces are equal and in- 
clined at an angle of 120°, their resultant is equal to 
either of them. Also, if three equal forces acting on a 
particle are inclined at angles of 120° to each other, the 
particle will be in equilibrium. 

(d ) y = 90° (Fig. 54). If y = 90°, or, in other 




words, the two forces are at right angles to each other, 
the general yalue of R becomes 

R* = P* + Q\ 



R = VP* + Q 2 

This result is derived immediately from the properties 
of a right-angled triangle, as are also the following rela- 
tions: 



cos a 



sin a 



P_ 
R' 

R' 



tana = 



P = R cos a; 
Q = R sin a; 
Q = P tan a. 



Also, sin a == cos /?, cos a = sin /?, and tan a = cot j3. 

132, Condition of Equilibrium for Three Forces acting 

GA a Particle, If three forces acting on a particle may 



132.] COMPOSITION OF FORCES. 137 

be represented by the sides of a triangle taken in order, 
the particle will be in equilibrium. This proposition is 
called the Triangle of Forces. 

Let P, Q, S, represented by A B, AC, AD respec- 
tively (Fig. 55), be three forces acting on a particle at A, 



and let abd be a triangle so drawn that its sides, taken 
in order, represent respectively the three forces; viz., 
ab represents P, bd represents Q, and da represents S; 
then is the particle A in equilibrium. 

Complete the parallelogram abdc. Since ac is equal 
and parallel to bd, the resultant of the forces ab, ac, will 
be the same as the resultant of P and Q; hence ad is 
the resultant of P and Q; but da, equal and opposite to 
ad, represents the third force S. Therefore, since this 
third force is equal and opposite to the resultant of the 
other two forces, the particle acted upon must be in 
equilibrium (125). 

The condition " taken in order" is essential and must 
be carefully noted. 

Cor, The converse of this principle is also true: that 



138 STATICS. |133. 

if three forces acting on a particle keep it in equilibrium, 
the sides of any triangle which are respectively parallel 
or perpendicular to them will be proportional to these 
forces. 

133. If three forces acting on a particle keep it in 
equilibrium, each force is proportional to the sine of the 
angle between the other two. 
. Let P, Q, 8, represented by AB, AC, AD respec- 




Fio. 56. 

tively (Fig. 56), be three forces acting on the particle A, 
and keeping it in equilibrium; then 

P : Q : S = sin CAD : sin BAD : sin BAC 

For take A b, Ac proportional to P and Q respectively, 
and complete the parallelogram Abdc, and draw Ad; Ad 
will be proportional to S and in the same straight line 
with it, since it has the direction of the resultant of P 
and Q, and is proportional to it. Then 

P :Q :S= Ab : Ac : dA; 

but, by trigonometry, 

Ab : Ac (= bd) : dA == sin Adb : sin dAb : sin dbA, 

= sin CAD : sin BAD : sin BAG. 



134.] COMPOSITION OF FOKCES. 139 

Therefore 

P : Q : S = sin CAD : sin BAD : sin 5.4(7. 

Cor. The converse of this proposition is also true, and 
gives a second condition of equilibrium for three forces 
acting on a particle, which may take the place of that in 
Art. 132 — viz. : If of three forces acting on a particle each 
is proportional to the sine of the angle between the direc- 
tions of the other two, the particle will be in equilibrium. 
The condition must be observed, however, that no one 




of the forces can fall between the directions of the other 
two; or, in other words, that the angular distance 
between no two of the forces can be greater than 180°. 
Thus the proportion may hold good for the three forces 
in Fig. 57, and also those in Fig. 58; but only in the 
former case is there equilibrium. 

134. Composition of more than Two Forces acting 
upon a Particle. The resultant of several forces acting 
in the same plane upon a particle may be found (Fig. 
59) by taking first the resultant of two of the forces; 



140 



STATICS. 



[134. 



then of this resultant and a third force; again, of the 
resultant of these three forces and a fourth force, and 
so on. 

This method may be most simply applied as follows: 

Let (Fig. 60) AB, A C, AD, 
AE represent the four forces 
P, Q, S, T, acting on the par- 
ticle at A. From the point a 
(Fig. 61) take ah inthedirec- 
e tion of P and proportional to 
it; then in the same manner 
take he to represent Q, cd to 
represent S, and de to repre- 
sent T. It is obvious (com- 
pare Fig. 59) that ac represents the resultant of ah, he, 
that is of P and Q; also ad of ac and cd, that is of P, 
Q, S; finally ae, the side which completes the polygon 




Fig 





dbede, is the resultant of ad and de, that is of the four 
given forces P, Q, 8, T. 

The numerical calculation of the magnitude and 
direction of the resultant in accordance with this con- 
struction, following the method already given (130), in- 
volves considerable labor. A more simple method is 
given in a subsequent article (140). 



136.] 



COMPOSITION" OF FORCES. 



141 



4* 



i 
k 



H 



135. Forces not in the same Plane. The method of 
finding the resultant of any number of forces acting on 
a particle, given in the first paragraph of the preceding 
article, is also applicable when the forces are not in the 
same plane. 

For example, let AB, AC, AD (Fig. 62) represent 
the three forces P, Q, 8 respect- 
ively, acting on the particle at A. 
The diagonal AE of the parallel- 
ogram ABEC will represent the 
resultant of the forces P and Q; 
also, if the parallelogram AEFD 
be constructed, the diagonal AF 
will represent the resultant of 
AE and AD; that is, of P, Q, and 
8. The figure thus constructed 
is sometimes called the Parallelopiped of Forces. 

If the forces are at right angles to each other, then 

AE 7 = AB 2 + AC = P 2 + Q 2 ; 
also, AF 2 = AE 2 + AD 2 ; 

.-. R 2 = P 2 -f <) a + 8\ 

If a is the angle between R and P, /3 between R and 
Q, y between R and 8, then 



Fig. 62. 



COS a 



-£> cos J3 



R' 



8 

cos r = ^ 



136. Condition of Equilibrium for more than Three 
Forces acting on a Particle. Any number of forces act- 
ing upon a particle will hold it in equilibrium, when 
they may be represented by the sides of a polygon taken 
in order. The forces P, Q, 8, T, and U (Fig. 63), 



142 



STATICS. 



[136. 



represented by AB, AC, AD, AE, AF respectively, 
will hold the particle A in equilibrium if they can 
be represented by the sides ab, be, cd, de, ea, taken in 
order of the polygon abede. For, supposing them to 
be so represented, it has been shown (134) that ae is 




Fig. 63. 

the resultant of the four forces P, Q, 8, T, and since 
the fifth force, U, represented by ea, is equal and 
opposite to ae, it must hold it in equilibrium ; that is, 
all the forces must be in equilibrium. This proposition 
is called the Polygon of Forces. It obviously applies as 
well to the case of forces not in the same plane. 

EXAMPLES. 

XIX. Parallelogram of Forces. Articles 126-136. 

[The forces are in all cases supposed to act on a particle, or at a 

point of a body (123).] 

1. Two forces, P= 7 lbs., $ = 24 lbs., act at right angles to 
each other: Required the magnitude and the direction of their 
resultant. 

2. Two forces, P = 13 lbs., Q = 7 lbs., act at an angle of 138°. 
Required the magnitude and the direction of the resultant. 

3. The force P=16 lbs. and the resultant i2=24 lbs., and 



137.] RESOLUTION OF FORCES. 143 

the angle between them is 42°: Required the other force, Q, and 
the angle between P and Q iy). 

4. The force Q = 5 lbs. and the resultant B = 6 lbs.; also, the 
angle between P and B (a) = 49° 30': What is the magnitude of 
Pand its direction? 

5. Of two forces, P= 16 and Q = 32 lbs. ; the angle between Q 
and B is 30° : Required B and the angle between P and Q. 

6. A peg in a wall is pulled by two strings with forces of 8 lbs. 
each; they are equally inclined downward (40°) to the vertical: 
What weight hung on the peg would give an equal strain? 

7. A peg in a wall is pulled by two strings, one horizontal with 
a tension of 21 lbs., and the other vertical with a tension of 28 
lbs.: What single force would exert an equal pull upon it? 

8. A weight is supported by two equal strings attached to nails 
in the ceiling and enclosing an angle of 60°; the tension of each 
string is 12 lbs.: What is the weight supported? 

9. Two forces in the ratio of 3 : 4, acting at right angles to each 
other, have a resultant 25: What are the forces? 

10. A boat is moored in a stream by two ropes attached to the 
shore making a right angle with each other; the tension of one 
(A) is 28 lbs., of the other (B) is 96 lbs.: (a) What is the actual 
force of the current, and (5) what angles do the ropes make with 
the direction of the current? 

11. In Fig. 48, P= 12 oz., Q = 15 oz.; the angle bad = 60°: 
Required W. 

12. Of two forces, P = 2 Q, and they act at an angle of 45°, and 
#=16: Find Pand Q. 

13. Three posts stand at the vertices of an equilateral triangle ; 
a rope is passed completely around them, the tension of which is 
24 lbs. : What is the pressure on each post? 

Resolution of Forces, 

137. Resolution of Forces. The process of finding 
the component forces whose combined effect shall be 
equivalent to a given single force is called the Resolution 
of Forces. It is the converse of the Composition of 
Forces. 



144 



STATICS. 



[138. 



To resolve a single force into two components, whose 
directions are given, all that is re- 
quired is to construct a parallelo- 
gram on those lines having the 
original force as the diagonal. 
Thus, if (Fig. 64) AC = R and 
the given directions are OX, OY, 
through C draw CD, CB parallel 
to the directions given; then AB, 

AD will be the components required. Similarly for any 

other directions. 
For example, let W (Fig. 65 or 66) be a weight hung 

by two strings knotted at a and attached at the points 

E and F. Produce a W vertically upward, and let ac 
1 




Fig. 64, 




Fig. 66. 



represent the weight W, and through c draw lines paral- 
lel to aE and aF respectively; then in the parallelogram 
so constructed ad, ad will be the conrponents of the force 
ac, which is equal and opposite to W. They give the 
tension of each string which supports the given weight. 

138. Rectangular Components. The case of the most 
importance in the resolution of a single force is that 
where the directions of the two components are at right 
angles to each other. The components in this case are 

(Fig- 67) 

AB — Bcos a, 



AD = R sin a 



138.] 



RESOLUTION OF FOECES. 



145 



Here a is tlie angle made by the direction of the first 
component with that of the resultant. If the body is 
free to move in one of these di- 
rections only, the component in 
this direction is called the effec- 
tive component, since this com- 
ponent alone influences the mo- 
tion of the body. — 

For example, suppose A to be 
a body either pushed along (Fig. 
68) on a perfectly smooth floor as 
with a rod, or pulled as by a string (Fig. 69), the force 
in each case acting obliquely, as CA (or A C). Then the 



Fig. 67. 



^dyS— 



~4 



.*• 



Fig. 68. 



yr 



Fig. 69. 



m. 



effective component, or that which alone influences the 
motion, is the one which acts in the direction of motion ; 
that is, BA or AB (=P cos fi). The other component 
produces no effect upon the motion. It has already been 
shown that if the surfaces are rough and friction has 
to be considered, the perpendicular component DA 
or AD (= P sin fi) in the one case increases and 
in the other diminishes the pressure on the surface, 
and so alters the resistance of friction (89). 

Again (Fig. 70), let a be a body resting on a smooth 
inclined plane ; the weight ( W) acts vertically down- 
ward (ac), but the body is obviously free to move only 
in the direction of the plane. The weight must hence 
be resolved along this line and along a line at right 



146 



STATICS. 



[139. 



angles to it ; thus, the components of the weight are ad 
and ab, of which ad is the 
effective component to produce 
motion, and (the plane being 
perfectly smooth) ab has no 
influence on the motion. Now, 
if HLK = lac = a, then ad = 
W sin a, and ab = W cos a. 




Fig. 70. 



?<& 



tzt 



Again, let a (Fig. 71) be a body of 
weight W rigidly attached to the 
point 0. In every position the 
weight acts vertically downward, 
but the body is free to move only 
in a direction perpendicular to the 
line of support. The two compo- 
nents of the weight in these direc- 
tions, as indicated in each case, are 
then ab — W cos a. and ad = W 
sin a, where a is the angle made 
with the vertical direction. 

The tension of the rod, or the pull 
or push on the point of support, is 
always equal to TTcos a (ab), and the 
effective component to produce mo- 
tion is always TTsin a (ad). Com- 
pare the different positions indi- 
cated, and note the values of the two 
components in each of them. In 
positions 2" and VII the moving FlG * "■ 

component is zero, and the tension is equal to W ; in position IV 
the moving component = W and the tension = 0. 

139. The explanation of the fact that a vessel may sail in a 
direction almost opposite to that from which the wind is blowing 
affords another illustration of this principle. Let AB (Fig. 72 or 
73) represent the direction of the wind. The resultant effect upon 
the sail MJSf may be represented by ab. This force is resolved 
into two components, one parallel to the sail (ac or db) and pro- 




140.] 



RESOLUTION OF FORCES. 



147 



during no effect, and the other perpendicular, ad. But the vessel 
is headed in the direction RH; hence to find the effective compo- 
nent of the wind in this direction the force ad must he again re- 
solved into the components of and ae. The tendency of of is to 
drift the vessel to leeward, and is nearly balanced by the resist- 
ance of the side of the vessel and keel (and the centre-board in 




Fig. 72. 



Fig. 73. 



the case of a sail-boat) against the water, and the component for- 
ward is ae. As a matter of fact there is always a little drifting, 
whence the motion of the boat is kept in the required direction 
by the rudder. The action of the rudder is itself another exam- 
ple of the same principle. It is seen in the figures that with the 
same wind two vessels may sail in exactly opposite directions. 

An explanation similar to the above may be applied to the 
motion of a windmill. 

140. Resolution of Forces along Two Axes at Right 
Angles to each other. The principle of the resolution of 
a force along two axes at right angles to each other may 
be conveniently employed to obtain the resultant of a 
number of forces acting at a common point. Let the 
forces P, Q, S, T (Fig. 74) be represented by AB, AC, 
AD, AB, and let X and Y be any two axes at right an- 



148 



STATICS. 



[140. 



gles to each other passing through A, The components 
of P, Q, 8, T are, geometrically, 

on the axis X Ab, —Ac, — Ad, Ae, and 

on the axis it. . . . Am, An, — Ar, — As. 

The minus sign indicates that the lines in question 
are measured — that is, that the forces act — in the oppo- 




Fig. 74. 

site direction to the others. The algebraic sum of each 
set of these components will give the components of the 
resultant along the respective axes. 

If a' , a" , a'", a iv are the angles which each of the 
forces makes with the axis X, all measured in the same 
direction as indicated in the figure (74), then the two 
sets of components will be : 

P cos cc' + Q cos a" + #cos a'" + Tcos a iY = x, 
and 

P sin a' + Q sin a" + S sin a"' + T sin a iv = y. 

The directions of x and y will be indicated by the al- 
gebraic signs belonging to the sum of the components in 



142.] 



EESOLUTION OF FOECES. 



149 



each case. If now the values of x and y be laid off 
from the point A (Fig. 75), we shall y 

have, by completing the parallelogram, 
the resultant (R) represented by AF, 
and 

R 

and 



= Vx* + 



tan 6 = S-; 

x 



% 



~x 



Fig. 75. 



so that the magnitude and direction of the resultant are 
determined. 

141. Condition of Equilibrium for Three or more 
Forces acting on a Particle. Any number of forces act- 
ing on a particle in the same plane are in equilibrium 
ivhen the algebraic sums of their components along any 
two axes at right angles to each other are equal to zero. 
For then R = 0, and this can only be true when 







and 



y = 



but x is the algebraic sum of the components along one 
axis X, and y along the other 
axis Y. 

This condition of equilibrium 
(analytical condition, it is called) 
may be taken in place of that 
given in Art. 13 G. 

of a Force 

A force may 
components 
along any three axes not in the 
same plane. For example, let AF (Fig. 76) be the given 
force, and let the three lines drawn through A represent 



142. Resolution 
along Three Axes. 

be resolved into 



B 



Fig. 76. 



150 STATICS. [142, 

the given axes. Then by reversing the construction of 
Art. 135 the figure represented in Fig. 76 is completed, 
in which AB, A G, AD are the required components. 

In the application of this method the axes are ordi- 
narily taken at right angles to each other. Thus, if a, 
f3, y are the angles between the given force (R) and the 
axes X (AB), Y (AC), and Z (AD) respectively, the 
three components x, y, z are 

x = R cos a, y = R cos /?, z = R cos y. 

The resultant of any number of forces, R u R 2 , R 3 , etc., acting 
in different planes at the same point, may be obtained by carrying 
out this method; for, take a x , fi x , y x to represent the angles made 
by Ri with the three axes X, Y, Z respectively, and a 2 , A, y* for 
the angles of the force JS 2 with the same axes, and so on; also, let 
x, y, z represent the sum of the components of the forces along 
the respective axes ; then 

x = R x cos cd -j- i? 2 cos a 2 -}- R 3 cos a 3 -j- etc., 
y = Ri cos fix -\- R 2 cos /3 2 -\- R 3 cos fi 3 -f- etc., 
z = Ri cos y x + Ri cos y<* -f- R 3 cos y 9 -f- etc., 
and 

Resultants V x* -f- f + z\ 

Also, any number of forces acting in different planes on a par- 
ticle will keep it in equilibrium if the sum of their components 
along any three axes at right angles to each other is equal to zero ; 
then 

x = 0, y = 0, z = Q. 

EXAMPLES. 

XX. Resolution of Forces. Articles 137, 138. 

1. A force of 150 lbs. is exerted in a due north-east direction: 
What portion of it is felt north? What portion east? 

2. A weight of 10 lbs. (Fig. 66, p. 144) is supported by two strings 
of equal length attached to nails in the ceiling: What is the tension 
of each of the strings for the following angles between them: 
0° (parallel), 30°, 60°, 90°, 120°, 150°, 180°? 



143.] RESOLUTION OF FORCES. 151 

3. A weight of 20 lbs. is supported by two strings at an angle 
of 140° ; one (a) goes (Fig. 65, p. 144) horizontally to the vertical 
wall, and the other (b) to the ceiling : What is the tension of the 
two strings? 

4. If the angle in example 3 is 150°, what are the tensions of a 
and 5? 

5. A picture, whose weight is 60 lbs., is supported by a cord 
attached to the upper corners and carried over a nail so as to 
include an angle of 80°. If the top of the picture is horizontal, 
what are the tensions of the strings? 

6. A horse drags a sled by a rope inclined at an angle of 15° 
with the ground; the tension of the rope is 600 lbs. : What is the 
effective component of the force exerted? What becomes of the 
other component? 

7. A weight of 18 lbs. is supported by two strings, one of which 
makes an angle of 30° with the vertical, and the other 60° : Find 
the tension of each string. 

XXI. Resolution of Forces along Two Rectangular Axes. 
Articles 140, 141 

1. Find the magnitude and direction of the resultant of the 
following forces: P=100 lbs., Q = 50, £=200; the angle be- 
tween P and Q = 60°, between Q and S = 120°. 

2. Required the magnitude and direction of the resultant of 
the following forces : P = Q = S= T — 100 lbs. The angles aTe 
as follows : between P and Q = 30°, between Q and S = 120°, 
between £and T= 30°. 

3. Required the magnitude and direction of the resultant of 
the following forces : P = Q = 100 lbs., S = T= 200 lbs. The 
angles are: between P and Q = 90°, Q and S= 135°, 8 and T= 90. 

4. Four forces, P, Q, 8, T, each equal in magnitude to 100 lbs., 
have the following directions : P= K 30° E., Q = K 30° W., 
8 = S. 60° W., T = S. 60° E. What is their resultant in direction 
and magnitude? 

5. Three forces, P, Q, S, each equal in magnitude to 200 lbs., 
act respectively K 45° E., and S. 45° E., and S. : What is the 
direction and magnitude of their resultant ? 

6. Three forces, P= Q = S — 100 lbs., act respectively E., 
N. 30° W., S. 30° W. : What force will hold them in equilibrium? 



152 



STATICS. 



[14a 



7. What force will balance the action of the four forces P = 
Q = 8= T= 100 lbs., acting respectively K 20° E., N. 70° W., 
S. 45° W., S. 45° E.? 

Composition and Resolution of Parallel Forces. • 

143. Parallel Forces. Forces are said to be pakallel 
when their lines of action are parallel. They are like 
parallel forces if they act in the same direction, and 
unlike when acting in opposite directions. 

144. (1) Like Paeallel Forces. The resultant of 
two like parallel forces acting on a rigid body is equal to 
their sum, acts in the same direction with them and at 
a point which divides the distance between them in the 
inverse ratio of the forces. 

Let (Fig. 77) the two like parallel forces P and Q act 
at the points A and B, supposed to be rigidly connected, 



g *' >iit 




>JC 



Fig. 77. 



and let them be represented by A H and B M respectively. 
Also, at A and B let two equal and opposite forces, 
s and s', be applied; they will not change the con- 



144.] PAEALLEL FOKCES. 153 

ditions, since they, taken alone, balance one another. 
Find the resultant AG of P and s, also the resultant 
BL of Q and s', and produce their lines of action till 
they meet at D. We may, by Art. 122, suppose them 
to act at D in their respective directions. Now resolve 
them (137) into their components again in directions par- 
allel to their original directions : the components s (Df) 
and s' (DJc) will balance each other and may be disre- 
garded; and the other components, P (Dh) and Q 
(Dm), both act in the line DC. Their resultant is 
therefore equal to their sum (R = P + Q), and may be 
regarded as acting at the point C. 

Again, since the triangles DC A, AUG are similar, 
and also the triangles DCB, BML, we have 



a) 

(2) 





DC AH 
AC~ GH~ 


P 
s' 


and 


DC BM 

CB ~ ML ~ 


s ' 


Then 


dividing (1) by (2), 

BC P 
AC~ Q m 





(3) 



This final equation proves that the line AB is divided 
at C into segments which are inversely as the forces. 
From (3), by inversion and composition, we obtain 

*° or 4g = Q (i) 



BC+AC P+Q' AB B 

n BG p BO P ,_. 

MS °' BC+AC = PTQ' ° r AB = R- < 5 > 



154 



STATICS. 



[145. 



145. (2) Unlike Parallel Forces. The resultant 
of two unlike parallel forces is equal to their difference, 
acts in the direction of the greater force, and at a point 
outside of it which divides the distance between the two 
forces externally in the inverse ratio of the forces. 

Let (Fig. 78) the two unlike parallel forces P and Q 
act at the points A and B, rigidly connected, and let 




Fig. 78. 

them be represented by AH and BM respectively. As 
before, apply two equal and opposite forces, s and s', at 
A and B. Find the resultant AG of P and s, and the 
resultant BL of Q and s'. Produce their lines of action 
till they meet at D ; they will meet in all cases unless P 
and Q are equal (150). Suppose the resultants to act at 
this point in their respective directions, and resolve them 
into components parallel to the directions of P (and Q) 
and s. Of these components s {Df) and s' {Dk) will 
balance each other, and Dh (= P) and Dm (= Q) will 
act at D in the same line and in opposite directions. 



145.] PAEALLEL FORCES. 155 

Their resultant will therefore be equal to their difference 
(or algebraic sum); that is, in this case 

R = F - Q. 

Also, this resultant may be regarded as acting at C in 
the line CD ; that is, parallel to and on the side of the 
greater force. 

Again, since the triangles ADC and A GF are similar, 
as also the triangles BDC and LMB ; then 

a) 



DC 
AC 


FG P 
~ AF~ s' 


DC 


BM Q 


BC 


~ ML~ s' 



Also, -UTr = -S7T- = T/- ( 2 ) 



Dividing (1) by (2), we have 

BC _P 
AG" Q' 



(3) 



That is, the point C divides the line AB externally into 
two segments which are inversely proportional to the 
forces. 

Also, we obtain from (3) 

(4) 



BO P 


BO P 
° r AB = R 


BG-AG~ P-Q? 


AG Q 


or AG -- Q 


BG-ACP-Q' 


0r AB~ K 



A1S0 ' BC=AC = T=Q> or AB = K (5) 

Taking together equations (3), (4), and (5) of the pre- 
ceding article, and also the corresponding ones of this 
article, it is seen that : Of two parallel forces and their 
resultant, each force is proportional to the distance be- 
tween the other two. 



156 



STATICS. 



[146. 



146. Experimental Verification. The principles dem- 
onstrated for like and nnlike parallel forces may also be 
verified by experiment. (1) Let (Fig. 79) xy be a rigid 
rod suspended at its middle point C. Also, let two 



a 

ft. 



x ^4 






o_ # y 



» S 






(? 



6o bji 



Fig. 79. 



Fig. 



weights P and Q be taken and hung on the bar ; they 
are then two like parallel forces. It will be found that 
in order to have equilibrium a weight R equal to P + Q 
must be hung by the thread over the pulley a, and also 
that P and Q must be so situated that 



P 

Q 



BC 
AC 



(2) Again, let (Fig. 80) Q be hung to the rod, and P 
suspended by the thread over the pulley ; they are then 
two unlike parallel forces. In this case, to maintain 
equilibrium R must be equal to P — Q. Also, Q and 
R must be so situated that 



£ 

R 



AC m 
BA ; 



xnau is. y^v — A J-*** 

Q AC 



147. In the case of more than two parallel forces the 
resultant is found as follows : First take the resultant of 
two of the forces, then that of this resultant and the 
third force, and so on ; the final resultant will be that 
of all the forces involved. 



149.] PARALLEL FORCES. 157 

The point at which this final resultant of several paral- 
lel forces acts is called the centre of parallel forces, 

148. Three Parallel Forces in Equilibrium. If three 
parallel forces keep a body in equilibrium, then each 
must be opposite to the resultant of the other two; that 
is, two of them must be like, and the third, equal to 
their sum, must act in an opposite direction at a point 
between them and at distances in inverse ratio to them. 

149. Resolution of Parallel Forces. A single force 
may also be resolved into components parallel to it and 
to each other. To accomplish this it is only necessary 
to remember the rule given that the distances from the 
resultant force to the components are inversely as these 
forces. 

For example, a weight W is hung at a certain point 
Con the rigid rod AB (Fig. 81); it is required to find 

~ a 



Fig. 8J. 

the component pressures P and Q at A and B respec- 
tively. Divide W into two such parts that P -f- Q = W, 

BC P 

and — ^ = -y. Or, from the final principle in Art. 145, 
AC (J 

, ., AB W -AB W 
take m = T , and ■ JS = -^ 

Again, ABG (Fig. 82) is a table with a triangular top; a weight 
Is placed at a point 0; it is required to find the pressure it exerts 
on each of the legs at A, B, and G. Draw AD, BE, CP, each 
through the point 0; then 

P BF P EG Q BG 

Q = AF> 8=AE aDd 8=DB^ (1) 



158 STATICS. [150. 

But since the triangles AFC, FBC, as also AFO, BFO, have the 
same altitudes, 

BF _ BFO _ BFO _ BOG 

AF ~ AFC ~ AFO ~ AOG 







P BOG 

' ' Q~ AOC 


Similarly, 
and 




P BOG 
8 ~ AOB' 

Q AOG 
8 ~ AOB' 


Therefore 


P: C 


) : 8= BOG : AOG : AOB. 



(2) 

As the problem would ordinarily be stated, the position of the 
point would give immediately the segments BF, AF, and BD, 




DC, etc., and therefore equation (1), remembering, also, that 
P-\- Q -\- 8 = W, would give the numerical solution. 

If the point is situated at the intersection of the three lines 
drawn from the vertices to the middle points of the opposite sides, 
then obviously 

P= Q = 8=iW. (See also 166, Cor. 1.) 

150. Couples. The case of two equal and unlike paral- 
lel forces is peculiar, since they have no resultant; in 
other words, their action cannot be balanced by the 
action of any single force. In Art. 145 above, if P === Q, 



150.] PARALLEL FOECES. 159 

then R = 0; but when R = 0, the values of BC (4) and 
AC (5) become infinite. This result may also be derived 
directly from Fig. 78, for as the difference between P and 
Q continually diminishes the point C recedes, and when 
P {AH) = Q (BM) the lines AG and BL will be 
parallel, and therefore C will be at an infinite distance. 

Two equal and unlike parallel forces are called a 
Couple. When acting on a free body a couple tends to 
produce rotation, and the body can only be kept in 
equilibrium by the action of a second couple whose 
moment of rotation (as defined below) is the same and 
in an opposite direction. 

The tendency of a couple to produce rotation is 
measured by the moment of the couple; this term will be 
further explained in the following article. This moment 
is equal to the product of either force into the distance 
between them. In the discussion of the general pro- 
blems which arise in higher Mechanics, couples play a 
very important part, but in this elementary discussion 
of the subject reference is seldom made to them, since 
problems involving the rotation of bodies are for the 
most part excluded. 

EXAMPLES. 

XXII. Parallel Forces. Articles 143-149. 

1. Find the resultant of the following parallel forces, and the 
position of the point at which it acts: (Compare Figs. 77 and 78.) 



(a) P = 5, 


Q = % 


AB = 4S. 


(b) P=12, 


Q = 1S, 


AG =81. 


(c) P=10, 


e = -4, 


^LB = 24. 


(d) P=14, 


Q = - 6, 


AG =■ 36. 



160 STATICS. [150. 

2 Find the force Q and the point at which it acts in the follow- 
ing cases- 

(a) P = 3, R = 8, AB = 40. 

(b) P = 5, .R = 14, 5(7 = 15. 

(c) P = 10, i2 = 6, ^5 = 24. 

(d) P=6, i?= 2, .4(7=48. 

3. A rigid rod, supported at the ends A and B, has a weight of 
48 lbs hung 6 feet from A and 18 feet from B: What pressures do 
the supports feel ? The weight of the rod itself is neglected here, 
as, too, in the following examples. 

4. ABC is a rigid rod ; at B a weight TTis hung, so that AB=12 
and BG =16; the pressure at A is 32 lbs. . What is the pressure at 
C, and what is W ? 

5. ABCD is a rigid rod; a weight of 4 lbs. is hung at the end 
A, and another of 6 lbs. at C (AC = 20 in.); it is supported at B 
(BA = 10 in.) and D (DA = 30 in.): What is the pressure on the 
supports ? 

6. A weight of 144 lbs is carried by means of a rigid rod on the 
shoulders (at the same height) of two men A and B; the distances 
from them are 5 and 7 feet respectively: What weight does each 
carry? 

7. A table has as its top an equilateral triangle ABC (Fig. 82) ; a 
weight ol 2C lbs. is placed at O, so that the perpendicular distance 
from O on BC = 18 in., and those on AC, AB each equal 36 in. -. 
What is the pressure on each of the three legs ? 

8. The top of a table is an isosceles triangle AB = A C = 2| feet ; 
at a point O, situated at a distance of 5 inches from each of the 
equal sides, a weight of 18 lbs. is placed: What pressure is felt at 
A, B, and C? (BAC = 90°.) 

9. A rod, whose weight acts at its middle point, rests on two 
vertical props placed at the ends. Where must a weight, equal to 
twice that of the rod, be placed that the pressure on the props 
shall be as 5 : 1 ? 

10. A rod, whose weight of 18 lbs. acts at its middle point, is 4 
feet long, and carries a weight of 90 lbs. 1 foot from one end : 
What are the pressures on two vertical props placed at the ends ? 



152.] MOMENTS. 161 



Forces tending to produce Rotation — Moments, 

151. In all the cases considered thus far, the tendency 
of forces to produce motion of translation has alone been 
involved (the remarks in regard to couples are to be ex- 
cepted). We have now to do with forces which tend to 
produce a motion of rotation. 

If a body has a fixed point or axis and a force acts on 
it in any direction except that passing through this point 
or axis, it will tend to produce rotation about it. This 
is seen when a force acts on the edge of a wheel free to 
turn on an axis. 

152. Moment. The moment of a force is the measure 
of the tendency of a force to produce rotation about a 
fixed point. 

The moment of a force with respect to any point may 
be demonstrated to be equal to the product of the force 
into the perpendicular distance from the point of rotation 
to the line of action of the force. 

This rotatory effect of the force consequently depends 
(1) on the magnitude of the force, and (2) on the per- 
pendicular distance from the fixed point upon its line of 
action — or, briefly, upon the length of its arm. 

For example, let (Fig. 83) a force P act at the point 
B on the rigid bar AB to produce rotation about the 
fixed point A : its moment is equal to the product of the 
force into its arm; viz., 

moment of P = P.AB. 

This moment is increased as the magnitude of P is in- 
creased, and also as the distance AB is increased. 
If the force acts obliquely, as in Figs. 84 and 85, in 



162 STATICS'. [153. 

this case also the product of the same factors gives the 
moment of the force, but the arm is now A G; that is, 

moment of P = P. AG. 

The same result would be obtained if the effective 
component of P were multiplied by the length of the 

Fig. 83. Fig. 84. Fig. 85. 

whole line AB. In the first case (Fig. 84), calling the 
angle ABC =■ /?, we have, as the moment of the force, 

ji e b P. AG = P.AB sin /?; (1) 



jg&..*j ) in the second case (Fig. 86), by resolv- 

\/' ing P, we have 

c 

Fig. 86. P sin fi.AB. (2) 

It is seen that (1) and (2) are identical. It is more con- 
venient, and less likely to lead to error, if the rule given 
on page 161 in italics is uniformly observed. 

153. Positive and Negative Moments. As one force 
may tend to turn a body in one direction, and another 
force in the opposite direction, it is necessary to distin- 
guish between their moments in this particular. This is 
accomplished by calling the moments positive (+) where 
the tendency is to turn the body in one direction, and 
those negative (— ) which have the reverse tendency. 

154. Geometrical Representation of the Moment of a 
Force. For purposes of demonstration it is often con- 



155.] 



MOMENTS. 



163 



venient to consider the moment of a force as repre- 
sented geometrically by double the area of a triangle 
having the line representing the 
force as its base and the given 
point as its vertex. Thus (Fig. 
87), the moment of the force P 

{AB) about the point C is equal L i ^^ 

to AB.CD, and this is double mo 87 

the area of the triangle ABC, 

155. The algebraic sum of the moments of two or more 
forces with respect to any point in their plane is equal 
to the moment of their resultant. 

Let AB, AD (Figs. 88, 89, 90) represent two forces, 
P and Q ; AC, the diagonal of the parallelogram A BCD 
constructed upon them, will be their resultant (R). The 
algebraic sum of the moments of AB (= AB.Eb) and 
AD (= AD. Ed), with respect to any point E, is equal 
to the moment of AC (= AC.Ec) with respect to the 
same point. There are three cases to be considered: 

(a) The point E falls without the angles DAB or 




DCB (Fig. 88), and hence the moments of P, Q, and R 
are all of the same kind. The triangle ABE is equal 
to the sum of the triangles ADC and EDC, for they 



164 STATICS. [155. 

have equal bases, AB and BC, and the altitude of the 
first triangle — that is, the perpendicular from E on AB 
— is equal to the sum of the altitudes of the other tri- 
angles; that is, the perpendiculars on the line DC from 
the vertices E and A, Hence 

triangle ABE = triangle ABC + triangle EBC, 

= triangle AEC — triangle ABE; 

.\ triangle ABE + triangle ABE = triangle AEC. 

If we multiply this equation by 2, we haye (by 1.54) 

moment of P + moment of Q = moment of R. 

(i) The point E falls within one of the angles named 
above (Fig. 89), and the moments of P and Q are of 




opposite kinds. The triangle AEB is equal to the 
difference of the triangles ABC, EBC, for they have 
equal bases AB and BC, and the altitude of the first 
triangle (the perpendicular from E on AB) is equal 
to the difference of the altitudes of the others (the per- 
pendiculars from A and E on BC). Hence 

triangle AEB — triangle ABC — triangle EB C, 
= triangle AEB + triangle A EC; 
,\ triangle AEB — triangle AEB == triangle AEC. 



156.] MOMENTS. 165 

Multiplying this equation by 2, we obtain (by 154) 

moment of P — moment of Q = moment of R. 

(c) The point E falls on the line of the resultant 
(Fig. 90). Since the perpendicular distances from B 



and D on AC are equal, the triangles A ED, AEB have 
the same base and equal altitudes, and are therefore 
equal. 

triangle A ED = triangle AEB, 

or triangle A ED — triangle AEB — 0. 

Multiplying by 2, we have 

moment of P — moment of Q = 0. 

This is in accordance with the proposition, for the 
moment of the resultant is obviously zero for this final 
case. The principle here established is an important 
one : the algebraic sum of the moments of two forces is 
zero for any point on the line of their resultant. 

The result reached m this article may be readily 
extended to any number of forces acting in the same 
plane, whether they intersect at a common point or are 
parallel. 

156. A tody, free to turn about a fixed axis and acted 
upon by forces in a plane perpendicular to this axis, will 



166 STATICS. [157. 

be in equilibrium, if the algebraic sum of the moments 
of all the forces about this axis is zero. 

According to the condition the body is only free to 
rotate in one plane perpendicular to its axis. In order 
that it should be in equilibrium the tendency to rotate 
in one direction must be balanced by the tendency to 
rotate in the opposite direction. This condition is satis- 
fied only when the algebraic sum of the moments of all 
the forces with respect to the axis is zero. By the con- 
cluding paragraph in the preceding article it is evident 
that the resultant (unless equal to zero) of all the forces 
must pass through this axis, for only in this case can its 
moment be equal to zero. 

This proposition is a most important one and has 
many applications; it is often called the Pkinciple oe 
the Levee. 

157. Free and Constrained Body. A body which may 
move unrestrained in any direction is said to be free. 
On the other hand, a body whose motion is restricted in 
any way is said to be constrained. 

An example of a constrained body is mentioned in 
the preceding article, and the condition of equilibrium 
for such a body, free to rotate only, is there given. 
Another example would be the case of a body strung on 
two wires and free only to slide in their direction; that 
is, to have motion of translation. The obvious condi- 
tion of equilibrium here is that the algebraic sum of the 
components of the forces taken in the given direction 
should be equal to zero. 

The conditions of equilibrium for a free body, acted 
upon by any number of forces in one plane, require that 
(a) it should not slide — that is, have motion of trans' 
lation —and that (b) it should not rotate. 



158.] CONDITIONS OF EQUILIBEIUM. 167 

For (a) the algebraic sum of the components in any 
two directions at right angles to each other must be 
equal to zero (141). 

For (b) the algebraic sum of the moments of the forces 
about any point in the plane must also reduce to zero 
(156). 

EXAMPLES. 

XXIII. Moments. Articles 151-156. 

1. A force, P = 12 lbs., acts at right angles to an arm 6 feet 
long: What is its moment? 

2. A rigid rod AB, 8 feet long and free to turn about B, is 
acted on by a force, P = 64 lbs., whose direction makes an angle 
of 40° with AB: What is the moment of P? 

3. A force, P= 150 lbs., acts at the extremity of a rod, AB, 
12 feet long, and at an angle of 160° : What is the moment of P 
about B ? 

4. A bar 6 feet long and pivoted at the middle has a weight 
of 24 lbs. hung at one extremity: What is the moment of the 
weight (a) when the bar is horizontal, (b) when it makes an angle 
of 40° below, and (c) of 60° above with the horizontal position? 

[Other examples involving the moments of forces are given 
under the Lever.] 

Summary of Conditions of Equilibrium. 

158. The various Conditions of Equilibkium for 
forces acting on a body in one plane, which hold true 
under the various circumstances, may be summed up 
here as follows : 

(A) For tivo forces: They must (1) act at the same 
point; (2) they must be opposite; and (3) they must be 
equal. 

(B) For three forces : They must, produced if neces- 
sary, (a) act at the same point, and have (a) the same 



168 STATICS. [158. 

or (/?) different lines of action ; or (b) they must be 
parallel. 

(a) a. If they act in the same line, their algebraic 
sum must be equal to zero (127). 

/?. (1) If they act at the same point and not in 
the same line, they may be represented by the sides 
of a triangle taken in order (132); or, (2) Each will 
be proportional to the sine of the angle between the 
directions of the other two (133, Cor.). 

(b) If parallel, two must be like parallel forces, 
and the third must be equal to their sum and act 
in an opposite direction to them at a point distant 
from them in the inverse ratio of the forces (148). 

(0). For more tlian three forces : 

(a) a. If they act in the same line, their algebraic 
sum must be equal to zero (127). 

/3. If they act at the same point (produced if 
necessary), but not in the same line, then: (1) They 
may be represented by the sides of a polygon taken 
in order (136) ; or, (2) The algebraic sum of their 
components along any two lines at right angles to 
each other must be equal to zero (141). 

(b) If they are parallel, the algebraic sum of their 
moments with respect to any point in the plane 
must be equal to zero. 

(c) If they act at different points or in different 
directions, then: (1) The algebraic sum of their 
components along any two lines at right angles to 
each other must be equal to zero; and also, (2) The 
algebraic sum of the moments of the forces about 
any point in the plane must be equal to zero (157). 



CHAPTER VII.— CENTRE OF GRAVITY. 



A. CENTRE OF GEAVITT OF BODIES — PLANE AND SOLID. 

159. Definition of the Centre of Gravity. The attrac- 
tion of the earth upon all particles of matter upon its 
surface is exerted in the direction of lines drawn to the 
centre. For the particles of the same body, or of neigh- 
boring bodies, these lines may be regarded as parallel. 
For a given body the resultant of all these parallel forces 
will act, whatever its position, at a certain point, called 
the centre of gravity. Hence 

Tlie centre of gravity of a body is that point at which 
the whole weight of the body may he considered as concen- 
trated; or — 

It is a point at which the body, if supported there and 
if acted upon only by gravity, will balance in every 
position. 

The definition may be extended to the case of a system 
of bodies if we suppose them and their centre of gravity 
to be rigidly connected. 

160. The Centre of Gravity of Two Bodies. Let P 

and Q (Fig. 91) be any two 
bodies of known weight. It 
is required to find the posi- 
tion of their centre of grav- 
ity. The weights may be 
considered as two like paral- 
lel forces whose resultant (144) 
will be equal to their sum and will act at a point which 




170 



STATICS. 



[161. 



shall divide the distance between them in the inverse 
ratio of the forces. Therefore, if the straight line AB 
be drawn and the point G taken on it, so that 

AG__Q 
BG " P' 

G will be the centre of gravity of P and Q. If this 
point be rigidly connected with the two bodies, the sys- 
tem, supported there, will balance in every position. 

161. The Centre of Gravity of any Number of Bodies. 

Let P, Q, S, and T (Fig. 92) be four bodies of known 
weight and occupying certain positions with refer- 
ence to each other. It is 
required to find their com- 
mon centre of gravity. On 
the straight line AB join- 
ing the positions of P and 
Q take E, so that 

AE_Q, 
EB P } 

then, by 160, E will be the 
centre of gravity of P and Q. Again, suppose P + Q 
to act at E, and on the line EC take F y so that 

EF _ S_ m 
EC ~ P+Q ; 

then F is the centre of gravity of P, Q, and 8. Again, 
suppose P + Q + 8 to act at F, and on the line .DP take 
G, so that 

FG _ T 

DG~ P+Q + S' 

then G is the centre of gravity of the four bodies P, Q, 




163.] 



CENTRE OF GEAVITY. 



171 



S, and T, and if it were rigidly connected with them the 
system would balance in every position. 

This method could obviously be extended, whatever 
the number or positions of the given bodies. 

162. The Centre of Gravity of a Straight Line. The 

centre of gravity of a straight line is at its middle point. 
Suppose the line to be made up of a series of material 
particles of equal weight ; the centre of gravity of each 
pair of them taken at equal distances from the centre of 
the line will be at this point. Hence the centre of grav- 
ity of the whole line will be at its centre. 

163. The Position of the Centre of Gravity of any Plane 
Figure determined by its Symmetry. The centre of gravi- 
ty of any geometrical figure, which is symmetrical with 





reference to an axis, lies in this axis. By a plane figure 
is here meant any material geometrical figure whose thick- 
ness is uniform and indefinitely small in reference to its 
other dimensions. 

Suppose the figure to be made up of parallel material 
lines, all bisected, according to the supposition, by the 
axis of symmetry. The centre of gravity of each of 



172 



STATICS. 



[164. 



these lines (162), and, therefore, of all of them taken 
together — that is, of the whole figure — will lie in this 
axis. If the figure has two axes of symmetry, their 
point of intersection will determine the centre of gravity. 
For example, the quadrilateral A BCD in Fig. 93, 
made up of two isosceles triangles placed base to base, is 
symmetrical with reference to the axis BD, for it bi- 
sects at right angles all lines drawn parallel to AC; 
hence the centre of gravity of the figure is in BD. So 
also if, in Fig. 94, AC and BD are both axes of sym- 
metry, the centre of gravity must lie at their point of 
intersection. 

164. Centre of Gravity of Regular Polygons. The posi- 
tion of the centre of gravity of the regular polygons is 
given immediately by this principle of symmetry. For 
example, in the equilateral triangle (Fig. 95) it is at G, 
the intersection of the three axes of symmetry drawn 




7^ 




Fig. 95. 



Fig. 96. 



Fig. 97. 



Fig. 98. 



from the vertices to the middle points of the opposite 
sides ; in the square (Fig. 96), at the intersection of 
the lines joining the middle points of the two opposite 
sides, or of the dotted lines joining the opposite angles ; 
in the regular pentagon (Fig. 97), at the common point 
of intersection of the five lines, each drawn from a ver- 
tex to the middle of the side opposite ; in the hexagon 
(Fig. 98), at the intersection of the three lines joining 
the middle points of the opposite sides, or of the three 



[165. CENTKE OF GBAVITY. 173 

dotted lines joining the opposite angles; and so on. In 
the circle, every diameter is an axis of symmetry ; the 
centre of gravity is consequently at the centre. 

165. Centre of Gravity of a Parallelogram. The cen- 
tre of gravity of a parallelogram is at the point of inter- 
section of the two diagonals. Let A BCD (Fig. 99) be a 
parallelogram whose diagonals intersect at G ; this point 
is the centre of gravity. Draw any line dgo parallel to 
the diagonal DGB; it is bisected by the other diagonal 
AGC. Tor from the similar triangles Adg, ADG, and 
Agl,AGB, 

DG~ AG' GB~ AG' 

dg go dg _ DG 

'''DG~1rB' ° r ~gb~ GB' 

But, by geometry, DG = GB ; hence dg = gd; therefore 
the centre of gravity of this line (162) must be at g. 
Hence, if the whole figure be considered as made up of 





material lines parallel to DB, the centre of gravity of 
each — that is, of the whole figure — must be in A C. In 
the same way it may be shown to lie in DB, and there- 
fore it must be at their point of intersection G. 

It may also be shown that the same point is deter- 



174 STATICS. [166. 

mined by the intersection of the lines (Fig. 100) EF 
and HK joining the middle points of the opposite 
sides. 

166. Centre of Gravity of a Triangle. The centre of 
gravity of any triangle is on 
the line drawn from either ver- 
tex to the middle point of the 
opposite side, and one third of 
the distance from this side. 

Let ABC be a triangle (Fig. 
101); from the vertex A draw 
AD to the middle point of the 
opposite side BC; also draw 
any line bdc parallel to BDC. 

Since the triangles Add, ABB, and Adc, ADC, are 

similar, 

bd __ Ad , dc Ad 

BD~AD f and DC~AD ; 




M 


dc 




Id BD 


BD 


~U~& 


or 


dc "~ DC 



But, by construction, BD = DC; hence bd — dc, and the 
centre of gravity of the line be is at d, on the line AD. 
Therefore it follows that the centre of gravity of all the 
material lines parallel to BC, of which the triangle may 
be considered as made up, lies in AD, and hence also 
that of the whole figure. 

Draw from the vertex B the line BE to the middle 
point E of the side AC; in the same way it may be 
proved that the centre of gravity of the triangle must 
lie in BE. Hence it must be at the intersection of AD 
and BE; that is, at G. 



167.] CENTKE OF GEAVITY. 175 

Connect DE; since the triangles AGB and DGE are 
similar, and also the triangles A B C and EDC, 

AG _AB AB_BC 

GD ~ BE 3 DE ~ DC' 

AG _BC _2 
""'■ GD~ DG~V 

That is, GD is one half of A G and one third of AD. 

Cor. 1. The centre of gravity of three heavy bodies 
of equal weight will coincide with the centre of gravity 
of the triangle whose vertices occupy the position of the 
three bodies. For (Fig. 102) the centre of gravity of 
the equal weights B and C will be at D, so that BD = 
DC (160). Also, the centre of gravity of B and C 
together at D, and of the third 
weight at A, will be at G, 

., , AG 2 AD 3 
S othat^- = I ,or- & - 3 = T 

But the same point G is also 
the centre of gravity of the 
triangle ABC. 

Cor. 2. The centre of gravity 
of a polygon can be found by dividing it into triangles, 
taking the centre of gravity of each by the above article 
and then proceeding as in Art. 161. The weights of 
the triangles are taken as proportional to their areas. 

167. The Centre of Gravity of a Solid Figure. The 

centre of gravity of a solid, which is symmetrical with 
reference to any plane, must lie in this plane. This fol- 
lows from the same consideration as that in Art. 163. 
If there are two planes of symmetry, the centre of gravi- 



j9 




176 STATICS. [168. 

ty will lie in their line of intersection ; and if three, at 
the point in which they all intersect. Therefore the 
centre of gravity of a sphere is at its centre ; of a cylin- 
der) at the middle point of its axis ; of a rectangular 
solid, at the point of intersection of three planes drawn par- 
allel to, and midway between, each pair of opposite sides. 

168. The Centre of Gravity of a Triangular Pyramid. 

Tlie centre of gravity of a triangular pyramid is on the 
line drawn from a vertex to the centre of gravity of the 
opposite side, and one fourth of the distance from that 
side. Let A BCD (Fig. 103) be a triangular pyramid. 
Take E, the middle point of D C, and draw BE; then F, 
one third of the distance on BE from E, is (166) the 



Fig. 103. 

centre of gravity of the base of the pyramid. Let bed 
be the triangle formed by the intersection of a plane 
drawn through any point b parallel to the base BCD. 
Draw AE meeting cd at e, and AF intersecting be at 
/; then it may be shown that / is the centre of gravity 



168.] CENTKE OF GEAVITY. 177 

of the triangle bed. For from the similar triangles Ace, 

ACE, aai&Aed, AED, 

ce Ae , ed _ Ae 

CE~AE } a ED~AE ; 

ce ed ce _ CE 

r 'CE~ED' ° r ~e~d~ED' 

But CE = ED; hence ce = ed, and e is the middle point 
of cd. Again, from the similar triangles Abf, ABE, 
and Afe, AFE, 

K.-AL and fL-Ai. 

BF~ AF a a EE-AF' 

K A- If _BF 

' "' BF~ Ftf ° r fe ~ FE' 

But BE = 2FE; hence If = 2fe, an&fe = \le, and / is 
a point on the line drawn from the yertex b to the mid- 
dle point of the opposite side one third of the distance 
from that side; hence (166) /is the centre of gravity of 
the triangle bed. 

If now the whole figure be thought of as made up of 
material triangles all parallel to BCD, the centre of 
gravity of each one, and hence of the whole pyramid, 
will lie in the line AF. For the same reason it will lie 
in the line BH, drawn from B to the centre of gravity 
of the side ACD; hence it will be at their point of in- 
tersection G. 

Since now the triangles HGF and B GA are similar, 
as also the triangles HFE and ABE, we have 
EG _FH FH _FE 

AG~ AB> AB~BE' 

EG _EE _1_ 
''' AG" BE~ 3* 



178 STATICS. [169. 

Therefore FG is one third of AG and one fourth of 
the whole line AF. 

Cor. The centre of gravity of four heavy bodies of 

" equal weight coincides with the centre of gravity of the 

triangular pyramid at whose vertices these bodies are 

situated. This follows in the same way as did Cor. 1, 

Art. 166. 

169. To find the centre of gravity of a pyramid, 
having any rectilinear polygon as its base: Divide this 
base into triangles by lines drawn from any angular 
point to the others, and suppose planes passed through 
the vertex and these lines. The pyramid is divided 
into a number of triangular pyramids. The centre of 
gravity of each of these will lie on the line drawn from 
the common vertex to that of its base arid one fourth 
of the distance from the base; therefore the centre of 
gravity of the whole pyramid will lie in a plane parallel 
to the base and one fourth the distance from it. 

Again, suppose the pyramid made up of similar poly- 
gons parallel to the base; the centre of gravity of each, 
and therefore of the whole figure, will lie on a line drawn 
from the vertex to the centre of gravity of the base 
(determined as in Cor.- 2, Art. 166). The centre of 
gravity of the pyramid will be at the point where this 
line intersects the plane above determined; that is, one 
fourth of the distance from the base. 

170. To find the centre of gravity of a cone: Suppose 
the cone be divided into an infinite number of triangular 
pyramids; then, by the reasoning of the preceding arti- 
cle, it is obvious that the centre of gravity must lie in a 
plane parallel to the base and one fourth the distance 
from it to the vertex, and also in the line joining the 



171.] 



CENTEE OF GEAVITY. 



179 



latter point with the centre of gravity of the base, and 
hence at their point of intersection. 

The centre of gravity of the material surface (taken 
in the same sense as in Art. 163) of a right cone lies on 
its axis and one third the distance from the base to the 
vertex. This is proved by showing it to be true first 
for a pyramid whose sides are triangles, and then pass- 
ing to the cone which is the limit of the pyramid when 
the sides are indefinitely increased in number. 

171. Problems. (1) Given the positions of the centres of 
gravity of two known parts of a body, to find the centre 
of gravity of the whole. Let the weights 
of the parts be w f and w" acting at the 
points g f and g"; then the centre of 
gravity of the whole will be on the line 
g'g" at a point G so situated that 



*£ 



W' 

w 1 '' 



g'G' 



c 



For example, suppose the parts to be 

the isosceles triangle ABO and the 

square BDEC, situated as in Fig. 104, 

and let AF= s (side of square). The 

centre of gravity of the triangle is at g' (g'F 

w' 1 
and of the square atg" (Fg" ="■£«); also, —, = - 





K 

F \ 




r 



Fig. 104. 



UF), 



There- 



at _1 _g"G 



fore -j-, = - = 
w" 

9"G = ^ 



~q ; but g'g" = | *, hence g'ff = is, and 



(2) Given the positions of the centres of gravity of a 
lody and of a known part, to find that of the remainder. 
Let W be the weight of the whole, and iv' of the part; 
then that of the remainder = W — w' (= w"); also, let 



180 STATICS. [171. 

the centre of gravity of W be at G, of w 9 at g 9 , and of 
the remainder {w") at g". Join g 9 and G, and take on 

the line produced ~-f = — n = -=- - -• then is q" the 

required point. 

For example, let the whole body be a circle ABC 
(Fig. 105) whose centre is G; and the part a second 




Fig. 105. 

circle whose centre is g 9 , and whose diameter is the 

w 9 1 
radius (R) of the larger circle. Then -= = — , and 

-77? 7 or — 7> = k-; a l so > q G — \R ; hence -^-y = — , 

Tf — w' V w") 3 ' ' * 2 ' Gg 9 S' 

and fly == i-jS. 

EXAMPLES. 

XXIV. Centre of Gravity. Articles 159-171. 

[The connecting rods mentioned are supposed to be rigid, and, 
except when otherwise stated, to be uniform and without 
weight.] 

1. Where is the centre of gravity of two bodies, A and B, 
weighing 4 and 5 lbs. respectively, rigidly connected by a weight- 
less rod 24 inches long? 

2. Three weights of 3, 6, and 12 lbs. are hung, at A, B, and G 



171.] CENTRE OF GRAVITY. 181 

respectively, on a rigid bar; AB = 6 inches and BG = 12 inches: 
Where will the bar balance? 

3. Four weights of 2, 3, 5, and 6 lbs., at A, B, G, and B, are 
connected rigidly in a straight line; AB = 10, BG = 8, GB = 18; 
Where is their centre of gravity? 

4. A rod AB, 18 inches long and weighing 4 ounces, has a 
weight P= 2 lbs. hung at the end B: Where will it balance? 

5. A rod AB, 24 inches long and weighing half a pound, has a 
weight P = 5 lbs. at a point 1 inch from B: Where will it 
balance? 

6. What weight must be hung at the end of a rod 3 feet long 
and weighing half a pound that it may balance 3 inches from 
that end? 

7. A rod 2 feet long and having a weight of 5 lbs. at one end 
balances at a point f of an inch from this end: What is its weight? 

8. A heavy rod, 24 inches long and weighing 3 lbs. , balances 
alone at a point 10 inches from one end: What weight must be 
hung at the other end in order that it may balance exactly in the 
middle? 

9. A ladder 40 feet long and weighing 60 lbs. has its centre of 
gravity 16 feet from the larger end: (a) If supported by two men, 
A and B, at the extremities, what will they carry? (b) Where 
should A stand to divide the weight equally with B1 

10. A uniform rod AB, 20 inches long and weighing 3 lbs., has 
a weight of 12 oz. at the end A, and one of 6 oz. two inches from 
B: Where will it balance? 

11. Weights of 8, 4, and 18 oz. respectively are placed at the 
vertices A, B, G of a triangle right-angled at B; AB = 18 
inches, BG = 9 inches : How far from G is their centre of gravity, 
and on what line? 

12. ABG is an isosceles triangle; AB = AG = 25 inches, BG = 
14 inches; weights of 4 lbs. each are placed at A, B, G respec- 
tively Where is their centre of gravity, measured from At 

13. ABC is a uniform rod bent at right angles at B; AB = 
BG =12 inches: Where is its centre of gravity, measured from Bt 

14. A uniform square board, ABGB {AB = 24 inches), weighs 
3 lbs. • (a) Where will it balance if a weight of 1 lb. is placed at 
At (b) if equal weights of 1 lb. each at A and Bl (c) if equal 
weights of 1 lb. each at A, B, and G ? 



182 STATICS. [172. 

15. Where is the centre of gravity of the remainder of a square 
board, ABCD (AB = 24 inches), (a) after a piece is cut out by- 
lines drawn from A and B to the centre? (b) Again, if a piece is 
cut out by lines joining the centre with the middle points of two 
adjacent sides? (c) Again, if one corner is cut off by a line join 
ing the middle points of two adjacent sides? 

16. ABC is an isosceles triangle; AB = AC = 20 inches, BC = 
32 inches; the upper portion is cut off by a line joining the 
centres of these sides : Where is the centre of gravity of the re- 
mainder? 

17. A circle having a diameter of 12 inches has a smaller circle 
cut out of it; the diameter of the latter is the radius of the former: 
Where is the centre of gravity of the remainder? 

18. A circle has a diameter of 16 inches; a smaller circle tan- 
gent to it and having a diameter of 12 inches is cut out of it: 
Where is the centre of gravity of the remainder? 

19. Find the centre of gravity of a frustum of a right cone 
whose altitude is 8 inches, and the diameters of the two bases 
6 and 12 inches respectively. 

20. Find the centre of gravity of a figure made up of two 
isosceles triangles (Fig. 93, p. 171): BE — 6, ED = 12. 

21. Find the centre of gravity of a figure made up of a square 
and an isosceles triangle, the latter having its base coincident with 
and equal to a side of the square (Fig. 104); the altitude of the 
triangle, 12 inches, is twice the side of the square. 

^22. Two uniform cylinders of equal lengths (= 20 inches), and 
having diameters of 12 and 6 inches, are joined so that their axes 
coincide: Where is the centre of gravity? 

B. APPLICATION OF THE PRINCIPLES OF THE CENTRE OF 
GRAVITY — EQUILIBRIUM AND STABILITY. 

172. Condition of Equilibrium. A body supported at 
a point, or on an axis, and free to turn about it, ivill be 
in equilibrium, under the action of gravity, if the verti- 
cal line through the centre of gravity passes through the 
point or axis of support. Such a body (Figs. 106, 107, 
108) is acted upon by two forces, (1) the weight acting 



[174. 



EQUILIBKIUM. 



183 



vertically downward through the centre of gravity, and 
(2) the reaction through the point or line of support. 
Therefore (158, A) the body can be in equilibriun only 
as these forces are equal and opposite; that is, the verti- 




Fig. 106. 



Fig. 108 r 



cal line through the centre of gravity must pass through 
the point of support. 

173. Hence, to find the centre of gravity of any body 
ly experiment: first support the body, as by a string, at 
one point, and when at rest extend the vertical line 
through the body; then suspend it from a second point, 
and also prolong this vertical line. The point of inter- 
section of these two lines will be the required centre of 
gravity; for, by the above article, the centre of gravity lies 
in each of these lines, and will therefore be at their point 
of intersection. This method is often useful in the case 
of irregular unsymmetrical bodies, to which the prin- 
ciples already given (163, 167) cannot be applied. 

174. If the vertical line through the centre of gravity 
does not pass through the point or line of support, the 
body will tend to rotate about this point or axis. For 
(Figs. 109, 110) the weight of the body ABC, represented 
by the vertical line Gd, may be resolved into two com- 
ponents, one, Ga, on the line GS drawn to the axis, and 
the other, Gb, at right angles to it. The first component 



184 



STATICS. 



[175. 



is balanced by the reaction of the point of support S, 
while the other component tends to make the body 
revolve. The same result would be obtained by taking 
the moment of the weight about S, which is equal to 





the product of the weight into the perpendicular dis- 
tance from S on to the line Gd. This moment vanishes 
(155, c) only when the condition in Art. 172 is fulfilled. 

175. Stable, Unstable, and Neutral Equilibrium. A 

body is said to be in stable equilibrium when, if slightly 
displaced, it tends to return to its original position; it is 




Fig. 111. 



in unstable equilibrium if it tends to move farther from 
its original position; it is in neutral equilibrium if, when 
moved slightly, it does not tend either to return or to 
move farther. 



175.] EQUILIBEIUM. 185 

Stable equilibrium is illustrated by Fig. 106, since, 
as seen in Fig. 109, a slight displacement is accompanied 
by a tendency to return to the original position. It is 
also shown in Fig. Ill, where the weights P, P bring the 
centre of gravity below the point of support. It is fur- 
ther illustrated by a compass-needle as usually supported, 
or by a hemisphere resting on the spherical surface, or 
a loaded circle (Fig. 112). In the last two cases it is to 
be noticed that the centre of gravity is above the point 
of support, and still the equilibrium is stable. 

Unstable equilibrium is illustrated by Fig. 107, since, 
as seen in Fig. 110, a displacement is accompanied by a 
tendency to rotate farther, the final position when the 
motion has been destroyed by the friction (108) being 
that of Fig. 106, or one of stable equilibrium. It is also 
illustrated by the case of a cone balanced on its apex, or 




c 
Fig. 112. Fig. 113. Fig. 114. 

the loaded circle in the position shown in Fig. 114. 

Neutral equilibrium is shown by Fig. 108, where the 
centre of gravity and point of support coincide. It is 
also illustrated by the case of a cone resting on its side, 
or of a sphere in any position on a horizontal surface; 
also by a wheel on an axle (see also Fig. 113). 

It will be seen that in stable equilibrium the centre of 
gravity is at the lowest possible point, and any change of 
position of the body raises it; it is on this account 



186 



STATICS. 



[176. 



that the body tends to return to its original position 
when displaced slightly. In unstable equilibrium it 
is at the highest possible point, and is lowered by a 
change of position. In neutral equilibrium it remains 
at a fixed distance from the support, whatever the posi- 
tion of the body. 

176. Stability of a Body resting on a Base. A body 
resting upon a base trill stand or fall according as the 
vertical line through the centre of gravity falls within or 
ivithout the base of support. By base of support is meant 
the salient polygon formed by lines joining the extreme 
points of support. For example, for a table with three 
legs it is a triangle formed by lines joining their extremi- 
ties. 

This principle is illustrated in Figs. 115, 116, 117, 




Fig. 115. 



Fig. 116. 



Fig. 118. 



118. In each case, let a vertical line through the centre 
of gravity ( G) represent the weight of the body; then, 
taking the moment (152) of the weight about the point 
C, which would be the axis in case of an overturn in 
that direction, the product of Wx EC measures the ten- 
dency of the body to retain its position (Figs. 115, 116), 
and the product Wx CE, Fig. 118, measures the tendency 
of the body to overturn. In Fig. 117 the line of the 
weight passes through the axis of rotation; hence its 



177.] 



STABILITY. 



187 



moment is zero, and the body is on the point of over- 
turning. 

In general, when the vertical line of the weight falls 
within the base, the product of the weight into the per- 
pendicular distance to the nearest side is called the mo- 
ment of stability. When the same line falls without, this 
product of the weight into the perpendicular distance 
to the nearest side is called the moment of instability. 

177. Conditions upon which the Stability of a Body 
depends. If, as in Figs. 119, 120, a force P acts, as indi- 
cated, at the point A, the body will be on the point of 
overturning when the moments of P and W about C are 
equal and opposite (156). If, in general, the arm of P 
is R (here BC), and of Wisr {EC), then 

PXE= Wxr. 

Also, if a force P acts to support a body tending to over- 
turn (as, for example, a prop), then the same equation 




1 -^JS 
Fig. 119. 



Fig. 120. 



Fig. 121. 



will hold true when the body is supported, as shown in 
Fig. 122, and the value of P given by the equation is 
the pressure on the prop. 

In the first case it is obvious that the greater the over- 
turning force required, the greater the stability of the 
body. But from the above equation, P must increase as 
W increases; that is — 



188 



STATICS. 



[177. 



(1) The stability is greater as the weight increases 
(other conditions being equal); e.g*, a stone tower is less 
easily blown over than a wooden one of the same dimen- 
sions. 

Also, P increases as r increases; that is — 

(2) The greater the base of support the greater the 
stability, other conditions remaining the same. Also, if 
the base is a regular polygon of given area, supposing 
the vertical line to fall through the centre, the stability 
is least if it is a triangle, it increases with the number 
of sides, and is greatest for the circle. 

Still again, if W and r are constant, P is increased as 
R diminishes, and conversely; in other words — 

(3) The greater the arm of the power the more readily 
is the body put on the point of overturning. For exam- 
ple, of two towers of different heights but otherwise 



P_Ail 




Fig. 133. 




alike, the higher would be the more easily blown over, 
since the resultant force of the wind would act at a 
greater distance from the base. 

Finally — (4) The stability is increased as the position 
of the centre of gravity is lowered. This relation does 
not follow from the equation, since that applies only to 
the case of equilibrium, where, as it was expressed, the 
body is on the point of overturning. 



177.] 



STABILITY. 



189 



In order that the body should actually be overturned 
P must continue to act (Figs. 123, 124, 125), diminish- 
ing continually as r diminishes, and becoming zero when 
the body is in the second position indicated in each 



&^- 




Fig. 125. 

figure. Hence work is done in accomplishing this re- 
sult, and this is estimated most simply by the product 
of the weight into the distance which the centre of 
gravity is raised (97). The work done increases as the 
position of the centre of gravity is lowered. For ex- 
ample, the initial values of P are the same in Figs. 123, 
124, and 125, as also the final values (= 0), but the arc 
Aa, through which P acts, and the height Fg, through 
which the weight is raised, are least for the highest 
position of the centre of gravity (Fig. 123), and greatest 
for its lowest position (Fig. 125). Thus, a stage-coach 
with a heavy load of trunks on top has its centre of 
gravity high, and is easily overturned by a slight irregu- 
larity in the road. 

EXAMPLES. 
XXV. Stability. Articles 172-177. 

[The centre of gravity in each case is assumed to be at the geo- 
metrical centre.] 
1. If the weight of the structure, of which ABCD (Fig. 115, 
p. 186) is a section, is 150 lbs., also AB=z 6 feet, AD = 10 feet: 



190 STATICS. [177. 

What force P will put it on the point of overturning (a) if acting 
at At (b) if acting at the centre of AD1 (c) If it rests on the side 
AD and the force acts at B and the middle point of AB respec- 
tively, what answers are obtained? 

2. In Fig. 122, p. 187, AD = AB = 10 feet, the angle ABC = 90° 
and BGE— 15°, the weight is 120 lbs. : What is the pressure on 
a prop placed (a) so as to act at B at right angles to BG ? (b) so as 
to act at the same point but standing vertically? 

3. A rectangular frame ABGD (Fig. 117, p. 186) is racked out 
of shape. If AB =12 feet, AD = 18 feet, What is the angle ADG 
when it is about to fall? 

4. A uniform stone tower, 8 feet in diameter, inclines 1 foot 
for every 10 feet of vertical height : What is the height of the top 
when it is about to fall? 

5. (a) A rough plane is inclined so that a cube resting on it is 
about to turn over, it not being able to slide: What is the angle? 
(b) What is the angle for an isosceles triangle (AB = AG— 10, 
BG — 16) if it rests on the side AB ? 

6. A table 6 feet square stands upon four legs, each of which is 
12 inches in from the adjacent edges; its height is 3 feet and its 
weight 24 lbs. : What is the least force required to put it on the 
point of overturning if applied at the edge (a) as a horizontal 
push? (b) as a pressure directly down? 

7. A table, having a circular top of 2 feet radius, is supported 
on three legs placed at the edge and at equal distances from one 
another; the height is 30 inches and the weight 20 lbs. : What is 
the least force that will put it on the point of overturning if applied 
at the top (a) as a horizontal push? (b) as a pressure down? (c) act- 
ing vertically upward? 

8. What work would be done in overturning a cylindrical 
column of stone weighing 40,000 lbs., 10 feet high and 4 feet 
diameter, supposing that the centre of gravity is on the axis (&; <,& 
the middle? ip) 1 foot from bottom? (c) 1 foot from top? 



CHAPTER VIII.— MACHINES. 

178. The machines are mechanical contrivances, by 
the use of which a force applied at one point is made to 
act at another with a change in either its direction or 
intensity, or in both. By means of them, for example, 
the power may raise a weight much larger than itself, 
or, on the other hand, it may give to the weight a 
velocity much greater than its own. In all cases, how-^ 
ever, the machine is only an instrument by which me- 
chanical energy is transformed; it never creates energy. 

179. The principle of the preceding article was laid 
down in Art. 98, where it was stated that, as follows 
from the law of the Conservation of Energy, in every 
case: "The work done by the power is equal to the 
work expended upon the weight." 

The work done by the force (P) acting is equal to the 
product of it (or its effective component, P cos 0*) into 
the distance (s) through which it acts; that is, 

P. s, or P cos /3.s. (1) 

The work done in raising the weight is equal to the pro- 

* When the force acts obliquely to the motion of the body, it is 
immaterial, in the estimation of the work done, whether the pro- 
duct of the effective component of the force into the whole distance 
is taken (= P cos fi.s), or the product of the whole force into the 
effective distance; that is, the resolved part of the motion in the di- 
rection of its own action (= P.s cos (5). 



192 STATICS. [180. 

duct of it (W) into the vertical distance (h) through 
which it is raised; that is, 

W.h. (2) 

If the work is done not against gravity in raising a 
weight, but against some other force producing a resis- 
tance, the work done is estimated by the resistance over- 
come (R) into the effective distance (d) ; that is, 

R.d. (3) 

It is, in general, found convenient to use the term weight 
as including the resistance, though the true distinction 
must not be forgotten. 

In every machine, according to the principle of work 
just stated, 

P.s = W.h, and P.s = E.d, (4) 

W = !> and -B = !- (5) 

The relation (5) may be expressed in this way, that: 
The Power is to the Weight (or Resistance) as the dis- 
tance through which the Weight is raised (or the Resist- 
ance is overcome) is to the distance through which the 
Power acts. 

180. Machines are then employed: (1) Where a small 
power is desired to raise a large weight or overcome a 
great resistance. In this case there is said to be a 
mechanical advantage ; but as seen from equation (4), 
if W is greater than P, s, the distance through which 
P acts, must be as many times greater than h, the dis- 
tance through which W rises. This is sometimes ex- 
pressed in this form: What is gained in power is lost in 
velocity. 



182.] MACHINES IN GENEKAL. 193 

Also — (2) Where an increased velocity is required; in 
this case h (or d) will be greater than s, but P must be 
as many times greater than W. There is then said to be 
a mechanical disadvantage, but, similar to the principle 
above, what is lost in power is gained in velocity. 

The cases where the attention is directed, in the use 
of the machine, solely to the diminished or increased 
velocity of the motion it transmits, the relation of P to 
W being overlooked, are obviously included in (1) or (2). 

(3) Machines are also occasionally employed where 
only change in direction is required, and here P = W (or 
R), and hence s = h (or d). 

181. The relation of P to W, established in equation 
(4), is that which is required in order that the power 
acting uniformly should raise the weight uniformly. If 
the value of the power were greater than that thus re- 
quired, accelerated motion would ensue; and if less, there 
would be retarded motion and the system would ulti- 
mately come to rest. 

The same relation of P and W will hold good if the 
system is at rest and the power simply supports the 
weight. The principles of statics make it possible to 

P 

deduce independently this ratio of -_ on the supposi- 
tion that the weight is at rest. In the pages which 
follow, the relation will be deduced by both methods: 
first in accordance with statics, and second on the prin- 
ciple of work. 

182. Virtual Velocities. In the second case given above, 
the principle of work may be stated in this form: 
If any machine, in equilibrium under the action of 
several forces, suffers a slight displacement consistent 



194 STATICS. [183. 

with the relations of the parts, then the algebraic sum 
of the work done by the forces will be zero, and con- 
versely. For such a case as this the velocities are 
imaginary, and are called virtual. This is sometimes 
spoken of as the principle of virtual velocities. This 
principle is essentially that involved in equation (4) or 
(5) of Art. 179. 

183. The Machines with Friction. In the statements 
which have been made in regard to the relation of the 
power and weight in the case of a machine, it has been 
assumed that the work done by the power was all ex- 
pended in raising the weight. In practice, however, 
there are various hurtful resistances to be overcome, 
chief among which is friction. Hence the work done 
by the power must always be greater than that which 
the equation (4) requires. For example, if ^represents 
the force of friction, and I the distance through which 
it is overcome, then the work done against friction, as 
shown in Art. 100, is F.l, and the equation must then 
be written 

P.s=.W.h + F.l. 

The law of the Conservation of Energy still holds good; 
but as the term F.l increases, the amount of work ex- 
pended in producing no useful effect, but merely use- 
less heat (110), is increased. Hence, although there is 
theoretically no limit to the mechanical advantage that 
may be attained (though always with a proportional loss 
of velocity) by an appropriately constructed machine or 
combination of machines, there is practically a limit; 
for, as the complexity increases, more and more of the 
power is expended without useful effect. 

To the resistance of friction must be added other re- 



185.] LEVEE. 195 

sistances which are additional drains upon the energy 
communicated to the machine, and which leave less to 
be expended in raising the weight. Among these are: 
adhesion of parts in contact, the stiffness of cords, resist- 
ance of the air, want of rigidity in the parts of the ma- 
chine. In the discussion in the following pages all these 
resistances are left out of account. The weights of the 
parts of the machines are also to be neglected unless 
otherwise stated. 

The modulus of a machine is the ratio of the amount 
of work practically done by it to that which theory re- 
quires. 

184. Simple Machines. The Simple Machines, or 
Mechanical Powers as they are sometimes called, are 
as follows: (1) The Lever, (2) Wheel and Axle, (3) 
Toothed Wheels, (4) Pulley, (5) Inclined Plane, (6) 
Wedge, (7) Screw. 

Of these machines the Wheel and Axle and Toothed 
Wheels are in fact modifications of the Lever. All of 
them involve the essential idea of a tendency to rotation 
about an axis, and hence to deduce the conditions of 
equilibrium for them the principle of the equality of mo- 
ments is employed (15 G). The Pulley is based upon the 
principle of reduplication, depending on the fact of equal 
transmission of force by a string ; or, in other words, 
that the tension of a rope at every point is the same 
(122). The Wedge and Screw are essentially modifica- 
tions of the Inclined Plane. 

I. Lever. - . 

A. General Principle of the Lever. 

185. The Lever in its simplest form is a rigid bar 
capable of being turned about a fixed axis called the/w/- 



196 STATICS. [186. 

crum, and supposed to be acted upon by forces in a plane 
at right angles to this axis. The bar may have any shape, 
straight, bent, or curved, and the directions in which the 
power and weight act may make any angles with it. 

186. For all forms of the lever the condition of equi- 
librium is that stated in Art. 156 for a constrained body 
only free to move about an axis in a plane at right angles 
to it. The power tends to produce rotation about the 
fulcrum in one direction, and the weight in the other. 
Hence — 

If the Power and the Weight are in equilibrium, the 
moment of the Power must be equal and opposite to the 
moment of the Weight. 

This may also be stated as follows : 

The Power is to the Weight as the perpendicular dis- 
tance from the fulcrum to the direction of the Weight 
is to the perpendicular distance from the fulcrum to the 
direction of the Power. 

For example, in Figs. 126, 127, 128, where the bar is 






B F S 



hur tew hw 

Fig. 126. Fig. 127. Fig. 128. 



* 



straight and the power and weight act at right angles 

to it, 

P.AF= W.BF, 

P_ BF 
or W ~ AF' 

The rule holds good equally well when the bar is not 
straight and the directions are oblique. The positions 



186.] 



LEVEE. 



197 



of the perpendicular distances from the fulcrum — that 
is, the arms of P and W — are to be carefully noted in 
the following figures, 129-134. For all of them the 
same equation holds good. 

The pressure on the fulcrum (the weight of the lever 
being neglected) in Figs. 126 and 129 is P -j- W, in Fig. 
127 it is W- P, in Fig. 128 it is P — W, and in the 
other figures it may be calculated by the parallelogram 
of forces. In the latter cases it is to be noted that, 
since the power, weight, and resistance of the fulcrum 
are in equilibrium, their lines of action produced must 
pass through the same point (158, B). 




Fig. 129, 



Fig. 131. 



In Fig. 132 the lever is curved like an iron pump- 
handle, the arms of the weight and power are the per- 



•*.•**•. 




Fig. 132. 



pendiculars BF and AF respectively, and the above 
equation is true: 

P.AF= W.BF. 



198 STATICS. [187. 

187. Three Kinds of Lever. The three forms in Figs. 
126, 127, 128 are sometimes called the three kinds of 
lever, though there is no essential difference between 
them. In the first kind the fulcrum is between the 
power and weight; if nearer to the latter, there is a 
mechanical advantage; if nearer to the power, a mechan- 
ical disadvantage. If the arms are equal, then P = W, 
as in the ordinary balance (191). 

In the second kind the fulcrum is at the end, and the 
weight nearer to it than is the power; in this case there 
is always a mechanical advantage. 

In the third kind the fulcrum is at the end, but the 
power is nearer to it than the weight, and there is there- 
fore a mechanical disadvantage. 

188. The first form of lever is illustrated by the crow- 
bar, by means of which, owing to the great difference in 
the lengths of the arms, a very great resistance can be 
overcome. Scissors and nippers are double levers of this 
class, and the handle and claw of a hammer form a 
curved lever. 

The distinction between the gain of power and loss of 
velocity, and the converse, as determined by the position 
of the fulcrum, is illustrated by the shears used by a 
tinman and a tailor respectively. Those of the former 
have short blades and long handles, and can consequently 
overcome a great resistance slowly; those of the tailor 
have short handles and long blades, and move quickly, 
so as to cut yielding materials. 

An example of a lever of the second class is a wheel- 
barrow : the fulcrum is at the centre, or axis, of the 
wheel, the weight acts down at the centre of gravity of 
the load and barrow together, and the power is applied 
at the handles. A nut-cracker or a lemon-squeezer is 
an example of a double lever of this kind. 



189.] 



LEVEE. 



199 



The human fore-arm is an example of a lever of the 
third class: the elbow- joint is the fulcrum, the weight is 
grasped in the hand, and the power is applied by a 
tendon from the muscle above attached very near the 
elbow, and acting obliquely. There is consequently a 
very serious mechanical disadvantage, but in its place is 
gained great rapidity of movement. A pair of tongs is 
another example of a lever of this kind. 

189. The following cases involve the principle of the lever. 
DF(Fig. 133 or 134) is a heavy rod hinged at F, so that it is free 
to turn in a vertical plaae, and supported either by a string carried 
from C up to E (Fig. 133), or by a prop from G to E below (Fig. 



J> c A 





2f C J3 



Fig. 133. 



Fig. 134. 



134). In this case the power (P) is the tension of the string (or 
thrust of the prop), and its moment is P.AF; the weight is that of 
the bar acting at its centre of gravity B, and its moment isW.BF. 
The value of P, derived from the equation 

PAF = W.BF, 



will give the tension of the string (or thrust of the prop) needed 
just to support the rod. 



200 STATICS. [190. 

190. The Lever on the Principle of Work. It has been 
shown in Art. 179 that, in the case of every machine, if 
friction and all other hurtful resistances are eliminated, 



P.5 = 


W.h, 


P 

w ~ 


h 

s' 



or -W= -„■ (1) 



Here s is the distance through which the power acts, 
and h the distance through which the weight is raised. 
If a resistance (R) is overcome through a distance d, 
then 

P.s = R.d, 

~R = 7- (2) 

By the use of these equations the relation of the Power 
to the Weight (or Resistance), when the Power raises 
the Weight uniformly, can be obtained. This value of 

P 

-r^ is the same as that deduced on condition of equilib- 
rium, on the principles of statics. 

In the case of the lever, suppose (1) that the power 




Pig. 135. 



always acts vertically (Fig. 135) and turns the lever 
from the position AB to that of A'B'\ then the effective 



190.] LEVEE. 201 

distance through which it acts is A'O (= s), and the 
height through which the weight is raised is B'D (= h). 
Therefore 

L. - KR - EE - ?I 

W ~ A'C ~ A'F " AF> 

.-. P.AF = W.BF, as in Art. 186. 

(2) Suppose the power and resistance to act continu- 
ally at right angles to the straight leyer AB (Fig. 136) 




Fig. 136. 

while it turns from the position AB to A f B\ Here the 
power acts through the arc A A' (= s), and the resist- 
ance is overcome through the distance of the arc BB' 
(= d). Hence 

_P BJ? _ BF 
R ~ AA' ~ AF ; 

.\ P.AF= R.BR 

In general, the relation may be obtained after the same 
manner, whatever the shape of the lever or the directions 
of P and W (or R). In each case, however, it must be 
remembered that s and h (or d) are not necessarily the 
actual, but always the effective, distances. 



202 STATICS. [191. 



B. Some Special Applications of the Principle of the 

Lever, 

I. BALANCE. 

191. The Balance is a contrivance used for measur- 
ing the mass of bodies ; or, in familiar language, of deter- 
mining their weight by comparison with that of certain 
assumed units. (See Arts. 54, 55.) In its ordinary form 
it consists of a beam, so constructed as to be at once strong, 
rigid, and light. This beam is poised on a knife-edge, in 
the middle, as a fulcrum, often resting on a plate of agate. 
From the extremities of the two equal arms are sus- 
pended pans of the same size and weight. The object 
weighed is placed in one pan, and the counterpoise is 
adjusted to balance it in the other. 

192. A good balance must satisfy these three condi- 
tions: it must be (1) true, (2) stable, and (3) sensible. 

(1) It is true when the arms are of exactly the same 
length and weight, and when the scale-pans are also just 
equal. It will then be rigidly true that P = W. If, 
however, the arm of the pan in which the object is 
weighed is longer, then a smaller amount of it will bal- 
ance the given counterpoise, and the purchaser in such a 
case would be defrauded, and conversely. This inequali- 
ty would be proved by exchanging the two objects. 
If the apparent weight in one pan is a, and in the 
other b, the true weight will be equal to V ab. 

(2) The balance must also be stable; that is, after being 
slightly disturbed it must return to its original position. 
In order to satisfy this condition the centre of gravity 
must be below the axis on which the beam turns, for if 
above it would be in unstable equilibrium, and if on the 



193.] 



BALANCE. 



203 



axis the equilibrium would be neutral; that is, the beam 
would balance in every position (175). 

(3) The balance must be sensible ; that is, when the 
weight in one pan slightly exceeds that in the other, 
this difference must be indicated by the inclined posi- 
tion of the beam when it comes to a state of rest. The 
sensibility is obviously greater as the angle of deflection 
increases for a constant difference of weight; this angle 
is often measured by a long slender rod which is at 
right angles to the beam and turns with it. The degree 
of sensibility to be attained in a given case depends 
upon the object for which the balance is to be used; for 
example, a balance suitable for use in chemical analyses 
should indicate distinctly a difference of ^ of a milli- 
gram; that is, yo-Jto" °f a gram. 

193. The conditions upon which the sensibility of a 
balance depends are: (1) the length of beam, (2) the 



jr. -£'- 

'* — Z? \ * 

■f£_ -A 

X 



Fig. 187. 



weight of the beam, and (3) the position of the centre 
of gravity. 

Let A and B (Fig. 137) be the points of the beam at 
which the scale-pans are hung, and let C be the axis or 



204 STATICS. [193. 

fulcrum; then the line A CB will join these three points. 
Suppose also that the shape of beam is such that its cen- 
tre of gravity is at G, at which point its weight ( Q) con- 
sequently acts. A'B' (Fig. 137) represents the inclined 
position which the line AB takes for a given difference 
of weight in the two pans of W—W. The angle of 
deflection of the rod FCF' (= a) is the same as that 
of the beam. It is obvious that, for a given value of 
W—W, the greater the angle a the greater the sensi- 
bility of the balance. Suppose the whole in equilib- 
rium; then, by taking the moments about C (156), 

W.A'K = W'.B'H + Q.DG. 

But since A'K' = B'H, 

(W-W').A'K=Q.DG. 

From the similar triangles GA'K, DGE, 

A'K DG A'K A'C 

or 



A'C ' QE' DG ~ GE' 

... (W-W')AC=Q.GE. 

But GE = CG tan a; hence 

(W-W')AC=Q.CGtma, 

(W-W').AC 
tan *~ Q.CG ' 

In this final equation A C is one half the length of the 
beam, and CG is the distance of its centre of gravity 
below the axis. Now, as has been stated, the sensibility 
of the balance increases as a increases, for a given value 
( W— W). It is obvious, from this equation, that tan a 
is increased (1) by making A C, the length of the beam, 



194.] STEELYARD. 205 

greater; also, (2) by diminishing Q, the weight of the 
beam; and finally, (3) by diminishing CG — that is, by 
bringing the centre of gravity as near as practicable to 
the axis. 

The most satisfactory result will be obtained by con- 
sidering these conditions together, since they depend 
upon one another, For example, if the length of the 
beam is increased, its weight must be also, in order that 
it still be rigid; again, although the sensibility increases 
as the distance of the centre of gravity below the axis is 
diminished, the motion of the beam, as it tends to come 
to a position of equilibrium, becomes more slow, so that 
there is also a practical limit in this direction. 

The equation shows that the difference in weight is 
proportional to tan a, and for very small angles it is 
proportional to the angle itself (a). 

II. STEELYAKD. 

194. Common Steelyard. In the Steelyard we have, 
in the place of the fixed arm and varying counterpoise 
of the ordinary balance, a varying lever-arm and a fixed 
counterpoise. The bar is made heavier at one extremity, 
and to this end is attached the hook or scale-pan; near 
it is the point of support. This axis is consequently 
near the centre of gravity of the whole, but usually does 
not coincide with it. In order to graduate the steelyard, 
it is necessary to determine first the zero-point of the 
scale, and then the distance to be marked off from it for 
each unit of weight (e.g. , 1 lb. ) and fraction of it. 

Let AB (Fig. 138) be the steelyard, supported at C. 
Represent the weight of the whole by Q acting at the 
centre of gravity G. In order that the bar should 
balance horizontally about C when there is no weight 



206 



STATICS. 



[194. 



on the hook, it is necessary to place the selected counter- 
poise P at such a point, D, that 

Q. CG = P. CD, 

This point D is then the zero of the scale, or the posi- 
tion of P for lbs. at A, 



M-il-G- 





Fig. 138. 



Fig. 139. 



Let now a weight W be placed on the hook so that 
it acts through A; then the counterpoise P will balance 
it at B, if (156) the moments about C vanish; that is, 

W.AC + QMC = P.CB; 

or, since Q.GG = P. CD, 

W.AC = P.GB - P. CD = P (CB - CD), 

or W.AC = P.DB, 

W.AG 



and 



DB 



If W — 1 lb., then the value of DB gives the position 
of the one-pound notch on the scale, and at twice this 
distance from D will be the two-pound notch, and so on. 
If, as in Fig. 139, the centre of gravity is on the other 
side of the fulcrum, the position of the zero-point D 
will also be changed, but the value of DB is obtained in 
essentially the same way. 



195.] 



STEELYAED. 



207 



195. A form of the steelyard as actually employed is 
seen in Fig. 140. It will be observed that both sides of 
the bar are graduated, and moreover there is a second 



d I 




Fig. 140 

ring to support it at C. When the steelyard is turned 
over and supported at C, the weight has a shorter lever- 
arm, and consequently, the counterpoise remaining the 
same, with this second graduation the 
instrument is adapted for heavier weights 
than in the first case. 

A common form of the steelyard is also 
seen in the post-office scales, where the 
object to be weighed is placed on a plat- 
form, and the counterpoise slides along 
the graduated arm. 

Another very simple form of balance 
involving the same idea of a varying 
lever-arm is seen in the contrivance often 
employed for weighing letters (Fig. 141). 
When there is no weight at B, the 
weight of the instrument acts through its centre of 




Fig. 141. 



208 STATICS. [196. 

gravity directly below the point of suspension C. If 
now a letter is placed between the springs at B the 
position is slightly changed, so that the moment of its 
weight is equal to the moment of the weight of the 
instrument in its new position. 

196. Danish Steelyard. In the Danish steelyard no 
counterpoise is employed, but the adjustment is made 
by shifting the position of the supporting-hook, and 
consequently giving the weight of the bar a longer or 
shorter lever-arm. Let (Fig. 142) AB represent the 



H 3 



Fio. 142. 

bar, heavier at the end B, and let its weight Q act at G. 
Suppose a weight Wto be hung on the hook at A; then, 
in case of equilibrium, we have 

Q.CG = W.AC; 
but CG = AG-AC 9 

.\ Q(AG-AC) = W.AC, or (Q + W)AC=Q.AG; 

Q.AG 



.: AC 



Q+W 



The arm is graduated by letting W = 1 lb., 2 lbs., etc., 
in succession; since Q and AG have constant values, AC 
is thus obtained for each case. 

197. Roberval's Balance. In many forms of balance 
in common use^ instead of two scale-pans suspended 



196.] 



STEELYAED. 



209 



HlMM 




Fig. 142a. 



from a beam above, there are two platforms supported 
from beneath, upon one of which is placed the object to 
be weighed, and upon the other the counterpoise; or (as 
in the post-office scales alluded to in Art. 195) there is 
one platform, and the place of the other is taken by a 
graduated arm upon which slides a constant counter- 
poise. In such balances it is P P 
essential that the indications , 
should be accurate, no mat_ 
ter what the position of the 
load on the platform. 

The way in which this end 
is often attained is illustrated 
by RobervaVs balance (Fig. 
142a) ; CD, EF are here two bars of equal length, piv- 
oted to the upright support at A and B; they are also 
jointed at C, E, and D, F, thus forming a rectangular 
frame. The equal pans are supported at H and K, 
and upon them respectively are placed the object to be 
weighed and the counterpose, as P, P at M and N. (In 
actual use the frame CDFE is generally concealed in the 
stand of the balance. ) 

In the figure it is seen that the weights P, P are at 
very unequal distances from the axis AB, but the ac- 
curacy is not impaired by this fact. To prove this, sup- 
pose two opposite forces, each equal to P, to act at K 9 
and two others similar at H\ they will not alter the 
previous conditions. We have now, in place of P at N, 
a force equal to P acting downward at K, and a couple 
(150) whose moment is P.KN; also, in place of P at 
M, we have an equal force acting downward at H, and a 
couple whose moment is P.HM. The two equal down- 
ward forces at H and K will obviously balance each 



210 



STATICS. 



other; the couples though unequal do not disturb the 
equilibrium, for they merely produce unequal strains at 
the fixed points A and B, and do not alter the effect of 
the other forces. 

ni. TOGGLE-JOINT. 

198. Toggle- Joint. Fig. 143 represents two combined 
levers, A B, BC, forming what is called a toggle- 
jo ikt. They are hinged together at B, forming an 
angle ABC = 2 a -; further, the lever AB turns freely at 
A, while the end of BC is free to move in the direc- 
tion xy, and acts against the resistance Q. Suppose the 




Fig. 143. 



force P to act vertically downward at B against the resis- 
tances R and B; if the system is in equilibrium, we have, 
by Art. 133, 



P_ 
R 



sin ABC sin 2a 2 sin a co^ a 



sin PBC sin a 



sin a 



2 cos a. 



But the effective resistance Q is only one component of 
R, the other being supplied by the reaction of the plane 
xy; hence 



Q = R sin a, and 

2 cos a 



P 

Q 



R 

2 



Q 



sin a' 



sin a 



tan a 



199.] 



TOGGLE-JOINT. 



211 



As now the levers AB, BC straighten out, the angle 
2a approaches 180°, and tan a approaches infinity as 
its limit; the mechanical advantage, therefore, when 
the levers are nearly in a straight line is enormously 
great, and hence a very great resistance can be over- 
come. The toggle-joint is seen in the arrangement by 
which the cover of a carriage is raised. 

199. The principle of the toggle-joint and its great efficiency 
are well illustrated by the Stone- Crusher invented by Mr. Blake, 
of New Haven. The accompanying cut (Fig. 144) gives a longi- 




Fig. 144. 



tudinal section of the machine. The parts show so clearly the 
relations explained in the preceding article that but little descrip- 
tion is needed. The power is applied to the wheel E, and as it 
revolves the central post alternately rises and falls. This serves 
to work the levers, or toggles, 0, 0. One end of them is station- 
ary on the right (corresponding to A in Fig. 143), and the other 
acts on the movable iron mass J swung on K. Finally this re- 



212 STATICS. [200. 

suits in the opening and shutting of the jaws P, P. The distance 
through which the movable jaw works is small, and the force 
exerted against any object placed between is enormous. In use 
the blocks of stone or ore are fed in from above, and the motion 
of the jaws rapidly crushes them down to uniform fragments of a 
size regulated by the distance between the jaws at the lowest points ; 
the fragments pass out by the shute A. Such a machine will 
yield about 10 tons of broken rock per hour. 

IV. COMPOUND LEVERS. 

200. It is seen in Art. 186 that in the lever, by making 
the arm of the weight yery short and that of the power 
very long, any required mechanical advantage may 
theoretically be obtained. In practice, however, vari- 
ous difficulties would obviously arise in an attempt to 
gain power in this way. To avoid them, and at the 
same time to have greater compactness, it is found 
better to employ a series of levers, in which the weight 
of the first becomes the power of the second, and so on. 

In Fig. 145, let AC, DF, GK be three levers, ar- 
ranged as just indicated. Q is at once the weight of the 



Fig. 145. 

first and the power of the second lever, and the same is 
true for S with reference to the second and third levers. 
Now if the whole is in equilibrium: 

P _ BO Q BF S _HK 

Q~BA 9 8~ ED 9 W ~ HO' 

By multiplying these ratios together, 

P_ BG_ EF_ KK 
W ~~ BA X £D X H9' 



201.] COMPOUND LEVEES. 213 

It is seen here that 

The final mechanical advantage is equal to the product 
of the several values for each successive lever. This 
principle is true not only for levers in combination, but 
in general for any compound machine. 

P 

In determining the ratio of -rr. for a compound ma- 
chine on the principle of work, it is to be noted that the 
essential point is to know the distances through which 
P and W act. If these can be determined, their in- 
verse ratio gives the ratio required, and the intermediate 
steps in the machine are of no importance. 

201. The principle of the compound levers finds an 
important application in the scales used for weighing 
very heavy objects, as, for example, hay-scales, or rail- 
road scales for cars loaded with coal (say 10 tons each). 
By a combination of a series of levers, and at the same 
time by the suitable distribution of the weight brought 
on the platform, this being supported at a number of 
points, very heavy weights may be determined with all 
desirable accuracy. The whole is balanced by a counter- 
poise, or series of them, used in connection with a steel- 
yard arm. 

EXAMPLES. 

XXVI. Lever. Articles 185-190. 

[The weight of the lever is to be neglected, except when otherwise 
stated.] 

1. The force P = 40 lbs. acts as in Fig. 126, p. 196; AF = 
8 feet and AB = 10 feet : What weight can be supported ? 

2. If (Fig. 127, p. 196) AB = 10, BF = 2, and the weight is 
120 lbs., what force Pis required to support it ? 

3. If (Fig. 128, p. 196) AB = 14, AF = 2, and P = 100 lbs., 
what weight can P support? 



214 STATICS. [201. 

4. What is the pressure on the fulcrum in each of the above 
cases ? 

5. AFG is a bent lever (Fig. 129, p. 197); AF = 14, FC = 16, 
AFC = 135°, P = 30 lbs. : What is W ? 

6. CFD is a bent lever (Fig. 131, p. 197); CF = 18, FD = 12, 
PD.F = 150°, PCW = 165°, and W = 60 lbs. : What is P ? 

7. If in Fig. 130, p. 197, GB = 16, BF = 2, and ACF = 80°, 
also P = 40 lbs., what is TT? 

8. What is the pressure on the fulcrum in examples 5 and 7 ? 

9. A heavy uniform rod BF (Fig. 133, p. 199), weighing 25 lbs 
and 2 feet long, is hinged at F; it is supported by a string carried 
from G(GF = 20 in.) to a point E, 12 inches vertically above F: 
What is the tension of the string ? 

10. A heavy uniform shelf BF(Fig. 134), 18 in. wide (= BF), 
weighing 36 lbs. , and hinged at F, is supported by a prop carried 
from C (CF = 12 in.) to a point E below F, so that GF — FE: 
What pressure does this prop feel ? 

11. A uniform stick 8 feet long, weighing 2 lbs. , is supported 
between the thumb and first finger; the one acts at the extremity 
as a fulcrum, and the other as a force at right angles an inch from 
it: What is the force required when the stick is horizontal ? when 
inclined 60° to the horizontal ? 

12. A rod weighing 10 lbs. has a weight of 10 lbs. at one end 
and of 20 lbs. at the other : Where must the fulcrum be in case of 
equilibrium ? 

13. Forces of 8 and 12 lbs. act at the extremities of a bar 16 feet 
long, and in directions making angles of 135° and 150° respectively 
with it : Where is the fulcrum in case of equilibrium ? 

XXVII. Balance. Articles 191-193. 

1. A body is equivalent to a weight of 12 lbs. in one pan of a 
false balance, and of 16£ lbs. in the other: What is the true 
weight ? 

2. A body is equivalent to a weight of 6 lbs. 4 oz. from one arm 
of a false balance, and of 4 lbs. 6 oz. from the other: What is the 
ratio of the lengths of the arms ? 

3. The true weight of a body is 15 oz., its apparent weight in 
one pan of a balance is 1 lb. : What would it seem to weigh in 
the other pan ? 



WHEEL AND AXLE. 215 



XXVIII. Steelyard. Articles 194-196. 
[The common steelyard (194) is intended unless otherwise stated.] 

1. The longer arm of a steelyard is 26 inches in length, the 
shorter 2§ inches ; the arrangement of the scale-pan (or hook) is 
such that P (= 2 lbs.) at B (Fig. 139, p. 206), if BG = 10 inches, 
balances 8 lbs. at A : (a) Where is the zero of the scale ? (b) If the 
■whole weighs 1| lbs. (= Q), where is the centre of gravity? (c) What 
must be the ^graduation for ounces ? (d) If P cannot be conveni- 
ently brought nearer than f of an inch to G, what are the greatest 
and least weights for which it can be used? 

2. The whole length of a steelyard is 24 inches; CO (Fig. 138) = 
i in., GA = £ in., P = 8 oz., and Q = 1 lb. : (a) Where is the 
zero of the scale ? (5) What is the length of graduation for 1 lb. ? 
(c) How large weights can it be used for ? 

3. In Fig. 140, p. 207, AG — 3 in. and AG' = 1 in. ; the counter- 
poise = 12 oz. : What is the length of a division on the scale for 
1 oz. in each position ? 

4. The weight of the beam of a steelyard is 3 lbs., and the 
distance of its centre of gravity is i inch from the fulcrum : Where 
must a counterpoise of 1 lb. 12 oz. be placed to balance it ? 

5. The length of a Danish steelyard is 30 in., its weight is 
4 lbs., and acts at a point 3 in. from one end; a body weighing 
12 lbs. hangs at the other end: Where is the fulcrum ? 

6. The length of a Danish steelyard is 28 in., its weight is 
3 lbs., acting at a point 4 in. from one end: (a) Where is the 1-lb 
notch? (b) the 2-lb.? 

II. Wheel and Axle. 

202. The Wheel and Axle, in its simplest form, 
consists of two cylinders of different sizes, rigidly con- 
nected and turning about a common axis; the larger is 
called the wheel, and the smaller the axle. The power 
is applied to the end of the rope wound about the wheel, 
and the weight is raised by a rope wound upon the axle. 
This is seen in Fig. 146. 



216 



STATICS. 



[203. 



203. Suppose the power and weight to act in the same 
plane perpendicular to the axis, as in Fig. 147. Treated 
in this way, it is essentially a form of lever, the fulcrum 
being at F; and in case of equilibrium the condition 
(156) will hold good that the algebraic sum of the mo- 




Fig. 146. 



Fig. 147. 



ments of P and W about the axis must be equal to zero. 
That is, 

P.AF = W.BR 

If AF = E, and BF = r, then 



P^ 

W 



r 
B ; 



or- 



The Power is to the Weight as the radius of the axle is 
to the radius of the wheel. 

204. Since the power, if applied by a rope as here, 
must always act at right angles to the radius of the 
wheel, the relation given above holds good whatever the 
direction of P. The pressure upon the axle, however, 
will vary: if P and W are parallel and in the same direc- 
tion, it is equal to their sum; if in opposite directions, 



207.] WHEEL AND AXLE. 217 

to their difference; and in other cases it is to be obtained 
by the parallelogram of forces. 

205. The thickness of the rope is here neglected. 
Strictly speaking, half of this thickness should be added 
to each of the radii. As the weight is raised, and its 
rope consequently wound up on the axle, thus increasing 
its radius, and also that of the power is unwound, thus 
diminishing its radius, the relation of P to W will con- 
tinually vary, and in general, for the case supposed, P 
must increase. 

206. The Wheel and Axle on the Principle of Work. 

Suppose the power to continue to act uniformly while 
the system makes one complete revolution; then (179) 
s = 27tfi, and h = 2tzt. Therefore 

P.2ttR = W.Zttt, 

or P.R = W.r; 

P r 
. \ r= = -=-, as in Art. 203. 

207. Applications of the Principle of the Wheel and 
Axle. The principle of the wheel and axle applies to 
many other cases besides that here described. For 
example, the same relation holds good for the common 
windlass, as where a bucket is raised from a well, the 
power acting at the end of a crank-arm (Fig. 148). It 
also applies to the steering-wheel of a ship, where the 
power is applied to the handles on the circumference; 
or to the capstan (Fig. 149), where the axis is vertical, 
the power acts on a handle, and the rope connected with 
the weight (or resistance) leaves the axle in a horizontal 
direction. 



218 



STATICS. 



[207. 



A good example of this principle is seen in the fusee 
(Fig. 150), which is applied to some watches and clocks. 




FlG.148. 



As the spring unwinds its force diminishes; but by means 
of the fusee it is made to act on a continually increasing 
lever-arm, and by a proper adjustment the moment, or 
turning power of the force, can be kept constant. 





Fig. 149. 



Fig. 150. 



The wheel of a vehicle is useful in reducing frictional 
resistance, as explained in Art. 86; in overcoming ob- 
stacles in the road it acts as a continuous lever, hence 
the advantage of a large wheel. 



t] 



WHEEL A^D AXLE. 



219 



208. Chinese Windlass. The combination of the 
wheel and axle seen in Fig. 151 is sometimes called the 
Chinese Windlass, or the differen- A 

tial windlass. There are here two 
axles of different sizes, and the 
arrangement is such that as the 
rope is wound up on the larger axle 
it is unwound on the smaller one, 
it passing under a movable pulley 
which supports the weight. The 
upward ascent of the weight is con- 
sequently very slow, but the me- 
chanical advantage very great. Let 
R be the radius of the crank-arm, 
and a and b respectively those of the larger and smaller 
axles. The tension of the rope is obviously -j- W. If the 
machine is in equilibrium, then, since the rope tends to 
turn the smaller axle in the same direction, and the 
larger in the opposite direction, to the power, by the 
equality of moments: 




Fig. 151. 



P.R + 


iw>z = 


W- 


P.R = 


W{a- 


-^ 


P 

W ~ 


a — 1) 
2R ' 





This result, also easily obtained by the principle of 
work (206), shows that the mechanical advantage be- 
comes very great as the difference between the size of 
the axles is diminished. 



220 STATICS. 

EXAMPLES. 
XXIX. Wheel and Axle. Articles 202-208. 

1. The radius of the axle is 2 inches, that of the wheel is 2£ 
feet, and the power acting is 80 lbs. : What weight is supported ? 

2. A horse exerting a force of 800 lbs. walks in a circle having 
a diameter of 18 feet and turns, by means of a lever-arm, a verti- 
cal post about which a rope is wound : If the diameter of the post 
is 8 inches, what resistance (e.g. , that of a building which is being 
moved) can the horse overcome ? 

3. Four men, each exerting a force of 60 lbs. acting on separate 
lever-arms, 4 feet long, turn a capstan ; the radius of the circle 
about which the rope is wound is 6 inches: What is the pull felt 
upon the anchor? 

4. A weight of 500 lbs. hangs by a rope 1 inch in thickness; 
r = 8 in., and B = 4 feet; the power acts on a lever-arm without 
a rope: What is P? 

5. A power of 12 lbs. balances a weight of 200 lbs. ; the radius 
of the axle is 3 inches: What is the diameter of the wheel ? 

6. In Fig. 148, B = 18 in., and the weight of 250 lbs. rises 2 feet 
while the power makes 5 revolutions: What is P? 

III. Toothed Wheels. 

209. A Toothed Wheel is a circular disc provided 
with teeth on the circumference; such a wheel turning 




Fig. 152. 



on one axis interlocks with a second turning on another 
axis (Fig. 152), and in this way the force applied at 



210,] 



TOOTHED WHEELS. 



221 



the first is communicated to the second. There may 
be a mechanical advantage with a corresponding loss of 
speed, or a gain in velocity and a consequent mechani- 
cal disadvantage. 

When the wheels are small, the teeth are nearly rect- 
angular in form. When the wheels are very large and 
great force is employed, the shape of the teeth is a 
matter of essential importance, in order that the loss of 
power arising from their mutual friction and resistance 




Fig. 153. 

may he a minimum. The detailed development of this 
subject belongs to applied mechanics. 

As seen in Fig. 152, when the wheels are turned, the 
teeth roll upon one another so that the points in which 
their mutual resistance is felt, due to the power and 
weight acting, lie very nearly at the touching points of 
the two circles drawn. These ideal circles are called the 
pitch-circles, and are concentric with the wheels them- 
selves respectively. 

210. Relation of P to W. In Fig. 153, let the power 
(P) act on the radius E; while the weight ( W), supported 



222 STATICS. [210. 

by the rope, acts on the radius r, they tend to turn their 
respective wheels in the direction of the arrows. The 
resistance between the two wheels is felt (209) in the 
line QQ' ', and if the system is in equilibrium, by the 
principle of the lever (156), the moment of Q' about A 
is equal and opposite to that of P, and of Q about B to 
that of W, Hence the relations: 





P.R = Q'.AG, 




W.r = Q.BC; 


is, 


P.R Q'.AG 
W.r T Q.BC' 




P.R AG 2ttAo 




' * W.r ~ BG " %nBG 



But since the number of teeth in the two wheels is pro- 
portional to their circumferences, if t = the number of 
teeth in the power-wheel, and T those in the weight- 
wheel, we have 

P.R _ £ 
W.r ~ T 
This may be stated: 

The moment of the Power is to the moment of the 
Weight as the number of teeth in the Power-ivheel is to 
the number of teeth in the Weight-wheel. 

The final equation above may be written: 

■L - £. v t 

W ~ R T' 

This is another application of the principle explained 
in Art. 200, since, in the case supposed, the machine is 
really compounded of the wheel and axle and the toothed 
wheels. 



212.] 



TOOTHED WHEELS. 



223 



211. Toothed Wheels on the Principle of Work. Sup- 
pose (Fig. 153) that the power continues to act through 
one circumference of its lever-arm, %nR (= s); if no 
toothed wheels intervened, the weight would rise through 
a distance equal to the circumference of the axle, 
27rr (= h). But the ratio of the distances through 
which the toothed wheels will turn is the inverse ratio of 
their number of teeth; that is, if the smaller wheel has 
20 and the larger 40 teeth, the former will revolve twice 
'40\ ,„.,-.,'. . ,T 



J , while the other turns once; or, in general, — . 
fore 



There- 



W 



2?rr 
2ttB 



X 



r 

E X T 9 



as above. 



212. Application of Toothed Wheels. An illustration 
of the use of this machine is seen in Tig. 154, to which 




Fig. 154. 



the final equation of Art. 210 applies. Severa* series or 
" trains" of toothed wheels are employed in derricks 



224 



STATICS. 



[213. 



and cranes for raising yery large weights. One of these, 
with three pairs of toothed wheels, is represented in 
Fig. 155. The relation of the power to the weight here, 
by the preceding principles, is: 



t> 



V 



P r_ £ , 

~W ~~ isT X T * T f X T" 

For example, if P = 10 lbs., the radius of the axle 
(r) = 3 inches, that of the crank-arm (R) = 2-j- feet; 

P if, also, the number of 
teeth in each of the 
smaller wheels (t, t', t") 
is 20, and of the larger 
wheels (T, T\ T") 120, 
then the weight which 
could be raised, all hurt- 
ful resistances being left 
out of account, would be 
21,600 lbs., or about 10 
tons. 

213. An excellent illustra- 
tion of the use of toothed 
wheels, where a "gam in 
power" is required, is af- 
forded by the "back gears" 
of a large turning-lathe. The 
arrangement is such that the 
cone-pulley F (Fig. 156), by which the power is applied, turns 
independently of the wheel L and of the spindle of the lathe, 
which last are attached together. The "back gears" are two 
toothed wheels attached to a common axis, and so placed behind 
the wheels G and L, respectively, that by a slight adjustment they 
can be, when required, put in gearing with G and L. When the 
simple motion of the lathe only is needed, a pin connects the 
cone-pulley F with L, and then the spindle is turned directly at a 




Fig. 155. 



216.] TOOTHED WHEELS. 225 

rate depending, as mentioned in Art. 216, on the pulleys over 
which the belt passes. If, however, a greater resistance is to be 
overcome, as when a large object is to be turned, the pin connect- 
ing F and L is taken out, and the " back gears" connected with G 
and L. The motion of the wheel G is then communicated to the 
larger wheel behind; this gives to the axis of the latter a speed 
reduced in the ratio of the number of teeth of the two. This 
axis turns the second small wheel behind L, and this gives motion, 
reduced in rate as before, to L and the spindle connected with it. 
If the number of teeth in the two pairs of wheels are respectively 
20 and 80, then the speed of the spindle is reduced by this con- 
trivance 16 times, and a corresponding mechanical advantage is 
gained. 

214. The rack and pinion consists of a straight bar 
in which teeth are cut, into which fits the toothed 
wheel, which is revolved by a handle or screw-head. 
By this means the bar, and that to which it is attached, 
is raised or depressed. This arrangement is often em- 
ployed in practice; for example, in moving the tube of 
a microscope up or down. 

215. Toothed wheels are extensively used in the 
works of a clock. The point practically considered is 
here the relative velocity to be given to the successive 
axes; the relation of P to Wis not taken into account. 

216. Use of Belts. In machines the motion of one 
axis is communicated to another by the use of belts, as 
well as by toothed wheels; there may or may not be a 
change of velocity. The use of the belt or strap depends 
on the friction of the surfaces in contact (85). 

The velocities of the two axes, assuming that the strap 
does not slip at all, are in the inverse ratio of the radii 
of the wheels, and the mechanical advantage is in the 
direct ratio, as was true of the toothed wheels. 

Thus, in the case of the cone-pulley {E, Fig. 156) on 



226 



STATICS. 



[217. 



the shaft communicating the power in the shop, and 
that (F) of the lathe below, the velocity of the axis of 
the lathe will be greatest, and the power of overcoming 




Fig. 156. 

resistance the least, when the belt passes over the largest 
wheel of the latter, and conversely. 

In general, according as the straps are or are not 
crossed, the motion of the second wheel is in the oppo- 
site or the same direction as that of the first. 

IV. Pullet. 

217. The Pulley consists of a circular wheel turning 
about an axis which is attached to a surrounding frame, 
called the block. About the circumference of the wheel, 
which is .grooved, passes a rope, and at one end of this 
the power acts. Sometimes two wheels are placed side 
by side, as in Pig. 157. 

The necessity of making the wheel turn on its axis 
arises from the friction, which is very much diminished 
in this way; except for this, fixed pegs would answer as 
well 



219.] 



PULLEY. 



227 



The efficiency of the pulley is based upon the prin- 
ciple (122) that the tension of a given string is the same 
at every point. 

218. Single Fixed Pulley. In the single fixed pulley 





Fig. 157. 



the Power is equal to the Weight. This relation follows 
immediately from the principle stated above, for (Fig. 
158) the tension of the rope on both 



sides of the wheel A must be the 
same, and to satisfy this condition we 
must have 

P = W. 



T 



s 



L 



mji 



\-P 



k r 



Fig. 158. 



There is therefore no mechanical ad- 
vantage in the use of the single fixed 
pulley, but it serves to change the 
direction of the force applied. The 
tension on the beam at a is equal to 2P. 

219. Single Movable Pulley with Parallel Strings. In 

the single movable pulley with parallel strings the Weight 
is twice the Power. The tension on both sides of the 
wheel A (Fig. 159) must, as before, be equal to P; 



228 



STATICS. 



[220. 



therefore W is supported by two upward forces, each 
equal to P, and 

W=2P, or P = iW. 

If, as in Fig. 160, the rope passes over a second fixed 





Fig. 159. 



Fig. 160. 



■fP 



pulley, no change is made by this (218), for it is still 
true that W = 2P. The tension at I is P, and at a 
(Fig. 160) it is 2P. 

220. Single Movable Pulley with Inclined Strings. It 
was assumed in the preceding 
article that both branches of the 
rope were parallel; if, however, 
they are inclined at an angle 2 a 
JP (Fig. 161), then 

W = 2P cos a. 

In Fig. 161 the tension of the 
rope on both sides of the pulley 
is, as before, P, but here W is 
supported not by two forces each 
equal to P, but only by their components acting verti- 
cally upward. Since (131, a) the vertical line bisects 




Fig. 161. 



221.] 



PULLEY, 



229 



the angle 2a, these components are equal, and each has 
the value P cos a; hence 



W = 2P cos a. 

There is evidently a mechanical disadvantage here as 
compared with the preceding case, for W = 2P only 
when a = 0°, the two strings being parallel, and P 
increases as a increases. When a = 60° (2a = 120°), 
then P = W; and when a = 90° (2a°= 180°) and the rope 
is horizontal, P = oo ; that is, if the 
rope were perfectly flexible, no finite 
force could draw it out horizontal. 

221. Combinations of Pulleys. 

First System. Fig. 162 represents 
what is called the first system of 
pulleys ; here W = 2 n P. The pul- 
ley A is supported by two forces^ 
each equal to P; these act on the 
string which passes under the pulley 
B, so that its tension is 2P. The 
pulley B is consequently supported 
by two forces, each equal to 2P. 
Again, the tension of the stria g pass- 
ing from B under C is 2*P; and O is supported by two 
equal forces, each equal to 2 3 P; similarly for D\ and 
finally, the pulley E is acted upon by two upward forces, 
each equal to 2 4 P. Hence 




or, in general^ 



W = 2 b P, 
W = 2 n P, 



where n is the number of the pulleys. 



230 



STATICS. 



[222. 



The beam supports at a a tension cf P, also %P at b, 
4P at c, 8P at d, and 16P at e. 

222. tfecoratf tfystem. TT = nP. Fig. 
<*^ 163 represents another system of pul- 
leys. As here but one string is involved, 
its tension throughout — that is, each of 
the six branches — is equal to P. The 
weight is therefore supported by six 
forces, each equal to P; that is, 




or, in general, 



W= 6P, 



W=nP, 



where n is the number of strings rising 
from the movable pulleys. 

This form of pulleys is the one which 
is generally employed, though in practice, 
as remarked in Art. 226, the wheels are 
placed side by side in two blocks. 

If in this arrangement the 
weight w of the movable block is 
taken into account, the relation is 
then 

W+w = nP. 

223. Third System. In the sys- 
tem of pulleys shown in Fig. 164, 
W= (2 n -1)P. 

The tension of the string on 
which the power acts, that is at a, 
is P; hence the pull on the wheel 
B is 2P, and this force is felt up- 
ward at b; still again, the pull on Cis 4P, or 2 2 P, and 



^ 


^ 




' V 


if 




dc 


t* 




I 


^W 



Fig. 164. 



.] 



PULLEY. 



231 



the same upward at c; also, on D it is 8P, and the 
same upward at d. That is, the weight is supported by 
the four forces ; viz., P + 2P + 2 2 P + 2 3 P = 15P. 
Hence 

TT = 15P = (2 4 - 1)P, 

or, in general, 

W = (2 n - i)P. 
The tension on the beam is equal to W -\- P, or 2 4 P. 

224. The following are other forms of pulleys, for 
which the relations of P to W can be established in the 
same manner as for those already explained. 

In Fig. 165, W = 4P. In Fig. 166, W = 5P cos a 
(the tendency of the horizontal component of P is also to 
be noted). In Fig. 167, W = 81P = 3*P. 




Fig. 167. 



225. The Pulley on the Principle of Work. 1. Single 
Fixed Pulley, Here (Fig. 168) the power and weight 
act through equal distances; that is, s = h, and there- 
fore W = P. 

2. Single Movable Pulley with Parallel Strings. In 



232 



STATICS. 



[175. 



this case (Figs. 169, 170) the distance through which P 
acts is twice the height through which the weight rises, 

.-. s = 2h, and W = 2P, or P = iW. 




Fig. 169. 



Fig. 170. 



■*FP 



3. Single Movable Pulley with Strings not Parallel, 
The effective component of the power is P cos a (Fig. 
171), and since s = 2h, as above, 

W = 2P cos a. 

4. First System of Pulleys. Suppose that the weight 
(Fig. 172) is raised through a 
height h, the wheel D is evidently 
raised 2h; the wheel C, 2*h; the 
wheel B, 2*h; the wheel A, %%-, 
consequently the power P must 
act through a distance s = 2 b h; 
that is, W.h = 27*. P, and W = 
2 5 .P; or, in general, 

W= 2 n .P. 

5. Second System of Pulleys. 
If the weight is raised (Fig. 173) a height h, each of 
the 6 strings must be shortened by an equal amount; 
consequently the distance s through which the power 




PULLEY. 



233 



must act is equal to 6h. Hence P.6h = W.h, and W 
= 6P; or, in general, 

W = nP. 

6. Third System of Pulleys. Suppose that the weight 
is raised (Fig. 174) a height A; then, since the pulley 
D is fixed, the pulley C will on this account move down 
a distance h; but, since the point c also rises a height h, 



^3 



*mm 




Fig. 172. 




the pulley B will move through a distance 2(h) + h. 
Similarly, the motion of A will be through a distance 
2(27*. + A) + A, and the distance (= s) for P = 
2(2*h + 2h + h) +h = 15, or (2* - 1)A. Therefore 
Wh = (2* - 1)/*.P, and W = (2* - 1)P; or, in gene- 
ral, 

TT = (2* - 1)P. 

226. Application of the Pulley. The second system 
of pulleys (222) is that which is most frequently em- 



234 



STATICS. 



[22ft 



ployed in practice. For the sake of compactness the 
wheels are arranged side by side in each of the two 

blocks, as shown in Fig. 175. The relation 

of P to W is the same as in Fig 163. 

The theoretical relation of P to W is not 
practically attained, for both friction and the 
stiffness of the rope are yery serious resist- 
ances to be overcome. For many purposes, 
notwithstanding this loss, the pulley is a 
most useful mechanical contrivance. It is 
jj> often used in connection with the wheel and 
axle or with toothed wheels, as in derricks 
and cranes. It plays an important part in 
the rigging of a ship. The single fixed 
pulley (218) is often employed where it is de- 
sired to change the direction of the force; 
e.g., in the case of a well. 

EXAMPLES. 
XXX. PuUey. Articles 217-226. 

1. A man weighing 150 lbs. sits on a platform 
suspended from a movable pulley (B, Fig. 160, p. 228), 
and raises himself by a rope passing over a fixed 
pulley (A): Supposing the strings all parallel, (a) 
what force does he exert? (b) What upward force 
is needed if the rope passes under a pulley fixed to 
the ground before coming to his hand? 

2. In a combination of pulleys, as in Fig. 162, W = 
1152 lbs. and P= 72 lbs. : How many pulleys are there ? 

3. In a combination of pulleys, as in Fig. 163, W = 
336 lbs. and P = 42 lbs.: How many movable pulleys are there? 

4 In a combination as in Fig. 164, W = 840 lbs., P = 56 lbs. : 
What is the number of pulleys? 

5. In the single movable pulley, P = 100 lbs. : Calculate the 
value of TTif 2a = 30°, = 60°, = 120°, = 150°, = 180°. 



Fig. 175. 



227.] 



INCLINED PLANE. 



235 



6. What force is needed to support 500 lbs. by the first system 
of pulleys, there being 4 in all? What is the force if each pulley 
weighs £ lb. ? 

7. Find P as in example 6, if the second system of pulleys is 
employed. 

8. Find P as in example 6, if the third system is used. 

9. What is the relation of P to TTin Fig. 165, if the weights of 
the movable pulleys are taken into account? 

10. What is the relation of P toWm Fig. 167, if the weights of 
the movable pulleys are considered? 

V. Inclined Plane. 

227. The Inclined Plane, considered as one of the 
simple machines, is a rigid plane inclined to the horizon 
at an angle a, and upon it a weight is supported by a 
power acting in some definite direction. If a section 
be made perpendicular to the plane, the figure below 
(176) is obtained. Here HL is the length (I) of the plane, 




HK its height (h), and LK its base (b). The three 
forces acting upon a body, and, as we suppose, holding it 
in equilibrium, are the weight (W) acting vertically 
downward, the power (P) acting at some angle ft with 
the plane, and the resistance of the plane acting at right 
angles to it; these forces are supposed to act in the same 
plane. 



236 



STATICS. 



228. Relation of P, W, and R. First Method, If the 
three forces P, W, R, acting together at 0, are in equi- 
librium, then (133) each force is proportional to the 
sine of the angle between the directions of the other 
two. That is, 

P : W : R = sin WOR : sin POR : sin WOP, 

= sin (180° - a) : sin (90° - fi) : sin (90° + a+fi), 

= sin a : cos fi : cos {a -\- J3). 

^sin** n _ Wcos(a-{-/3) 



Hence 



P = 



cos/? ' 



R = 



cos /3 



Second Method. The above relation may also be ob- 
tained as follows: Since the forces P, W, R are in equi- 
librium (141), the algebraic sum of their components 
along any two lines at right angles to each other will be 
equal to zero. 

Take as these directions (Fig. 177) a line parallel to 
the length of the plane, and one perpendicular to it 




coinciding with the direction of R. Then, taken geo- 
metrically, the component of R along HL = 0, of P = 
ab t of W = — ad; also, along the other axis the compo- 



229.] INCLINED PLANE. 237 

nents of R, P, and W are respectively R, ac, — ae. Ex- 
pressing these conditions trigonometrically, we have, first, 

P cos /3 — W sin a = 0, 

D TF sin a 

or P = ^-; (1) 

cos p w 

and second, R -f- -P sin /? — TF cos # = 0, 

or P = W cos ^ — P sin /?. 

Substituting the value of P from (1) in the preceding 

equation, we obtain 

„ w W sin a sin 3 

R = W cos a — ^ — — , 

cos p 

PT(cos a cos P — sin a sin/5) m 
~~ cos"/? ; 

*«» (« + /») 

COS /? w 

229. Special Cases. The values of P and P in terms of 
W and the angles a and /?, derived in Art. 228, apply to 
all cases, whatever the direction of P. If now the power 
acts along the plane, or horizontally, these general equa- 
tions take a special form applicable to the particular case. 

(a) The poiver acts along the plane (Fig. 178). Here 
/3 = 0, and cos J3 = 1; hence, from the general value 

n IT sin a ... 

P = -7, — , we obtain 

cos (5 

P = W sin a, (3) 

and, from the general value of 

R = ' ■ ■, we obtain 

cos/? 

R = Wcosa. (4) 



238 



STATICS. 



[230. 



(b) The poiver acts horizontally (Fig, 179). Here 
fi = — a, cos (— a) = cos a. Hence, for 

W sin a . 
F = — , is obtained 



cos/? 



and, for 



It 



TTcos (<*+/?) 
cosytf 

i2 



Tf sin or _ 
cos or 

, is obtained 
W 



W tan a, 



(5) 



cos or 



= W sec ar. 



(6) 



From (5), if a = 90°, P — oo ; that is, no finite force 
can support a body against a vertical surface if the sur- 
faces in contact are perfectly smooth and there is no 
adhesion. This is only a special case of the general 
principle that the action of a force does not affect the 
motion of a body in a direction at right angles to that in 
which it acts. 

230. The results in (a) and (b) of the preceding 




Fig. 178. 



article can also be obtained independently by another 
method. 

(a) The power acts along the plane. Let the lines 
P, R, W represent the three forces holding the body at 



.J 



INCLINED PLANE. 



239 



A in equilibrium. From P draw BC parallel to the 
direction of W\ then the triangle ABC has its three 
sides respectively parallel to the three forces, and hence 
(132. Cor.) these sides are proportional to them. Again, 
the triangles ABC and KHL are mutually equiangular 
and similar, hence 



or 



and 



P\ 


W: 


R 


= AB 


:BC: 


AC, 








= HK 


\HL: 


LK; 




P 
W 


= 


HK 


sin a, 






R 
W 


= 


LK 

HL ~ 


cos a. 





(1) 



(2) 



The result in (1) is sometimes stated in this form: 
WJien the Power acts along the plane, the Power is to the 
Weight as the height of the plane is to the length. 

(b) The power acts horizontally. Let (Fig. 179) the 




Fig. 179. 



three forces P, R, W be represented by the three lines 
meeting at A. Through B draw BC parallel to the di- 
rection of W, and produce R to meet BC at C. The 
sides of the triangle ABC are respectively parallel to the 
three forces P, R f W, and therefore (132, Cor.) are pro- 



240 



STATICS. 



[231, 



portional to them; moreover, AB C and HKL are similar. 
Hence 

P: W : R = AB : BC \ AC = HK : KL : HL. 



or = = tan a, (3) 



p 
w 


= 


HK 
KL 


= 


tan a 


R 
W 


= 


HL 
KL 


= 


sec a. 



w 



The result in (3) may be stated as follows: When the 
Power acts horizontally, the Power is to the Weight as the 
height of the plane is to the base. 

231. Inclined Plane on the Principle of Work. Let 

(Fig. 180) the power P } acting at an angle (J3) with the 




Fig. 180. 



plane, raise the weight ( W) from L to H. The effective 
component of the force, parallel to the plane, is P cos /?, 
and the distance through which it acts is the length of 
the plane (l) ; therefore the work done by the power is 
P cos /3.1, that done upon the weight is Wh. Hence 

P cos fi.l = Wh, but h = I sin a; 

.\ P cos fi =Wsina. 



232.] INCLINED PLANE. 241 

(a) The power acts along the plane. For this case we 
have 

P.l = W.h, or P.l = W.l sin a, 

and P= W sin a. 

(b) The power acts horizontally. The effective com- 
ponent of P is P cos ( — a) = P cos a; hence, as before, 

P cos a.l = W.h = W.l sin a, 
P cos a = W sin <*, P = TT tan <*. 

232. Applications of the Inclined Plane. An applica- 
tion of the inclined plane is seen in the arrangement by 
which boxes or barrels are pushed up from the ground 
into a wagon. A carriage-road leading gradually up a 
mountain, or a railroad on an up-grade, are other ex- 
amples. If the power acts parallel to the plane, as is 
generally the case, it has to support only one component 
of the weight ( W sin a) at any one time (neglecting fric- 
tion). It is evident, however, that the intervention of 
the inclined plane does not diminish the amount of work 
to be done, as that is the same for a given vertical 
height, whatever the angle of inclination. 

The wedge (233) is sometimes considered as a com* 
bination of two inclined planes, base to base. The screw 
may be considered as an inclined plane wound around a 
cylinder. 

EXAMPLES. 

XXXI. Inclined Plane. Articles 227-232. 

[The plane is supposed to be perfectly smooth.] 

1. The angle of the plane is 20°, the weight is 120 lbs. : What 
force is required to support the weight (a) acting parallel to the 



242 STATICS. [232. 

plane? (b) acting horizontally? (c) acting at an angle of 30° with 
the plane? 

2. What is the reaction of the plane in the three cases in ex- 
ample 1? 

3. A force of 100 lbs. acts parallel to an inclined, plane: What 
weight can it support in the following cases — the angle of the 
plane is (a) 10°, (b) 30°, (c) 45°, (d) 60°, (e) 80°, (/) 90°? 

4. A weight of 100 lbs. rests on an inclined plane : What force 
acting parallel to the plane is required to support it, if the angle 
of the plane has the same values as in example 3? 

5. A force of 100 lbs. acts horizontally to an inclined plane: 
What weight can it support in the different cases given in ex- 
ample 3? 

6. A weight of 100 lbs. rests on an inclined plane : What force, 
acting horizontally, is required to support it in the several cases 
of example 3? 

7. A railroad has a grade of 88 feet to the mile: What force 
must the locomotive exert to support the weight of the whole 
train, taking that at 25 tons? 

8. The length, height, and base of a plane are in the ratio of 
5:3:4. Into what two parts may a weight of 56 lbs. be divided 
so that one part resting on the plane may be supported by the 
other hanging over the top vertically downward? 

9. If a horse can raise 600 lbs. vertically, what weight can he 
raise on a railway having a grade of 3° ? 

10. The grade of a railway is 44 feet to the mile: What power 
(acting parallel) is required to support any given weight? 

11. A body is supported on an inclined plane by a force of 40 
lbs. acting parallel to the plane ; but if the force acts horizontally 
it must equal 50 lbs. : Required the weight of the body, and the 
inclination of the plane. 

12. Two inclined planes of lengths 40 and 60 feet are placed so 
that they slope in opposite directions, and their equal heights, 12 
feet each, coincide ; a weight of 8 lbs. is supported in the longer 
plane by a string, parallel to the plane, passing over a pulley at 
the top, and attached to a second weight on the shorter plane: 
What is the second weight? 

13. Weights of 8 and 12 lbs. are supported in equilibrium on 
two inclined planes, so placed that their equal heights of 6 feet 



234.] 



WEDGE. 



243 



each coincide ; they are attached to the extremities of the same 
string passing, parallel to the planes, over a pulley at the top : 
What are the lengths of the planes, the angle of the first plane 
being 30°? 

VI. Wedge. 

233. The Wedge in its simplest form is a five-sided 
solid, of which two adjacent sides meeting in the edge 
are rectangles, the two opposite ends are triangles, and 
the back is a rectangle. The power is supposed to act 
in a direction perpendicular to the back, and, assuming 
the surfaces in contact to be perfectly smooth, the resist- 
ances are felt in the same plane perpendicular to the 
sides. 

234. Suppose the triangle in Pig. 181 to represent the 
section made by a perpendicular plane through the 




Fig. 181. 



wedge; then, if the three forces P, Q, R hold the wedge 
in equilibrium, their lines of direction will, if produced, 
all meet at some point 0. Also, since these forces are 
by supposition at right angles to the sides of the triangle 



244 



STATICS. 



[235. 



ABC, these sides will be respectively proportional to 
them; that is, 

P :Q :R=AB : AC : BO. 

If (as in Fig. 181) the section is an isosceles triangle, 
AC= BC, and Q = R; hence 



P_ 
R 



AB 

AC 



But AB = 2AD; and if ACB = 2a, AD = AC sin a, 
or AB = 2 AC. sin a; that is, 



^P 
i? 



2 A C sin a 
Z<7 



= 2 sin or. 



It appears from this equation that the mechanical ad- 
vantage increases as the angle of the wedge diminishes. 

235. Wedge on the Principle of Work. Let ABC 
(Fig. 182) be an isosceles wedge, and let the power act- • 

ing against the resistances Q and R 
(= 2R) force the wedge uniformly 
in a distance equal to DC. The 
work done by P is P. DC. The 
effective distance through which 
the resistance has been overcome is 
D'E. Therefore 

P.DC = 2R.D'E, 
D'E = D'C sin or; 
.-. P.DC = 2R.DCsma, 
Fig. 183. or P = 2R sin a. 

236. Application of the Wedge. In practice the rela- 
tion established for the wedge has little value, for the 




236.] WEDGE. 245 

resistance due to friction is enormous. The wedge, 
however, is an important instrument; it appears in many 
cutting tools, such as the knife, chisel, axe, plane, and 
so on. It should be noticed that in the case of the plane, 
for example, if used for cutting soft wood, the angle is 
small and the edge sharp; for harder wood the angle is 
larger. The tool for planing iron has a very large 
angle — varying, say, from 60° to 80°. 

When the wedge is employed as for cleaving wood, the 
resistance due to the cohesion and friction combined is so 
great that, instead of the pressure supposed in the above 
article, a blow from a heavy body, as an axe, is used to 
drive it in. In this the principle explained in Art. 105 is 
employed; the energy of a heavy body in motion being 
expended through a very small distance, and hence over- 
coming a great resistance. A nail is a familiar form of 
wedge, and its use further illustrates this principle ; it 
depends upon friction for its hold in the substance into 
which it has been driven. 

EXAMPLES. 
XXXII. Wedge. Articles 233-235. 

1. A wedge is isosceles in shape and has an angle of 20° ; if 
P = 40 lbs., what is the resistance on each face? 

2. A wedge is isosceles, and the angle 90° ; a force of 100 lbs. 
acts at the back: What are the other two forces ? 

3. A wedge is isosceles, the power acting on the back is 40 lbs., 
and the forces on the other sides are 60 lbs. each: What is the 
angle of the wedge? 

4. A wedge is isosceles and has an angle of 60° : What is the 
relation between the three forces? 

5. The wedge is right-angled and the three sides have lengths 
of 15, 12, and 9 (back) : If P = 100, what are the other two forces? 

6. The angle of the wedge is 30°, the back is 10, and one side is 
20: What is the ratio of the three forces? 



246 STATICS. [237. 



VII. SCREW. 

237. The Screw consists of a solid cylinder with, a 
raised portion passing spirally about it, which is called 
the thread. This thread is either rectangular or tri- 
angular in shape. It may be regarded as generated by 
the revolution of a rectangle, in the one case, or an 
isosceles triangle, in the other, about the cylinder at the 
same time that it advances uniformly parallel to the 
axis, and at such a rate that in each revolution it goes a 
distance equal to its own width. The two kinds of 
threads are shown in Figs. 183, 184. 

The screw in use works in a nut whose parts are com- 





Fig. 18a. Fig. 181 

plementary to those of the screw, so that the one fits 
closely into the other. The power acts at the end of a 
lever-arm to turn the screw in the nut; either one may 
be made stationary, so that the other moves with refer- 
ence to it. The pressure of the weight or resistance is 
felt in the direction of the longer axis of the screw, but 
the resistance between the screw and nut is felt at each 
point of contact between them and perpendicular to 
their common surface. 

238. In the screw the Power is to the Weight as the 
distance between the threads is to the circumference 
described by the Power. 

Suppose the surface of the cylinder, about which the 



238.] 



SCEEW. 



247 




thread passes as a spiral line (its thickness being neglect- 
ed), to be unrolled on a plane. A rect- 
angle with a series of equal triangles 
is the result, as shown in Fig. 185. 
Here AB is the circumference of the 
cylinder = 2/tt, if r is the radius of 
the cylinder; GAB is the angle between 
the thread and a horizontal line, called 
the angle of the screw; BG is equal to 
the distance between the threads, also 
called the pitch of the screw. 

BG - AB tan a; 

. \ distance between threads = 27tr tan a. 

Again, at each point of contact between the screw and 
the nut a force acts horizontally 
and holds both (1) P, acting on the 
lever-arm R (Fig. 186), and (2) W, 
acting vertically downward in equi- 
librium. Let the sum of these 
partial horizontal forces be repre- 
sented by F; then, taking moments 
about the axis, 

P.R = Fr. (1) 

For P and F tend to turn the 
cylinder in opposite directions. 
Also, on the principle of the in- 
clined plane, since the force F 
acting horizontally [229 (5)] sup- 
ports the total weight W, 

F= JT. tan a. (2) 




Fig. 186. 



18 


! 


STATICS. 


rom (1) and (2) 


P.R 


= W.r tan a, 


ultiplying by 2zr 


P 

W 

> 
P 

w 


r tan a 
~ R > 

%nr tan a 
%nR ' 



But 27tr tan a = distance between the threads, and 
27rR is the circumference described by the power-arm; 
hence the relation already given is deduced. 

239. Screw on the Principle of Work. Let the power 
acting on its lever-arm cause it to describe a complete 
circumference %nR\ the work done by the power is then 
P.%7tR, At the same time the weight has been raised 
(or resistance overcome) through a distance equal to 
that between two consecutive threads. Therefore 

P.%nR = Wx dist. between threads, 

P __ dist. between threads 
W ~ 2^rR '' 

240. Application of the Screw. The screw is practi- 
cally a most important mechanical instrument, being 
used in many cases where a great weight is to be raised 
or a heavy pressure to be exerted. For example, build- 
ings are often raised by the combined use of a number 
of screws, and screw-presses are employed for many dif- 
ferent purposes. 

By increasing the length of the power-arm, or dimin- 
ishing the distance between the threads, any required 
mechanical advantage may theoretically be obtained ; in 
fact, however, a limit is soon reached; friction is a 



242.] SCEEW. 249 

serious element, and the modulus of the machine is 
small. The use of the common screw, as, for example, 
in binding two boards together, depends for its efficiency 
entirely upon friction. 

241. Micrometer Screw. The screw is also employed 
for measuring very small distances, and is then called a 
micrometer screw. In this case the relation in Telocity 
of motion of the parts is the matter considered, and the 
relation of P to TTis lost sight of. The principle of the 
micrometer screw will be clear from the following 
remarks. Suppose a screw with 100 threads to the 
inch: it is obvious that each complete revolution will 
advance the screw if the nut is stationary, or the nut if 
the screw is held firm, through a distance of -^ of an 
inch. Suppose, further, that the head of the screw 
is a circle whose circumference is graduated into 100 
equal parts: then, if it is arranged with a fixed index, it 
is easy to turn the head — that is, the screw — through y^- 
of a revolution, and this will cause an advance of -^ of 
yj-g- of an inch; that is, 10 ooo of an inch for the screw 
itself. Screws with very fine threads, and hence giving 
very slow motion, find many applications in physical 
apparatus. 

242. Differential Screw. The differential screw gives 
a greater mechanical advantage, and hence a slower 
motion (which may be the end desired), than can be 
conveniently obtained from the simple form. Here a 
larger screw turns in a fixed nut, and a smaller One with 
a less pitch turns inside of it. The power acts directly 
on the larger screw, and the resistance is felt by the 
smaller. Hence it is evident that while the large screw 
descends the small screw ascends, and the actual motion 



250 



STATICS. 



[243. 



of the platform is the resultant of these two opposite 
velocities. 

The relation of P to W is given immediately by the 
principle of work. Suppose the power to act through 
one circumference (%nR); the larger screw will descend 
a distance equal to the distance between its threads ( p), 
and the smaller screw will ascend the distance between 
its threads ( p') ; hence the difference of these two will 
represent the distance through which weight is raised 
(or resistance moved). 

P_ h _ p-p' 
W ~ s " %tzR ' 

The Power is to the Weight as the difference of the 
distances between the threads of the two screws is to the 
circumference described by the power-arm, 

243. Endless Screw. Fig. 187 represents what is called 




Fig. 187 



an endless screw. Here there is a cylinder with a uni- 
form thread fitting into the teeth of a toothed wheel. 
The number of threads in the screw of the cylinder A B 



243.] SCREW. 251 

will determine the number of teeth of the wheel 0, 
which will be advanced in one revolution of the former. 
Upon this depends the mechanical advantage of the 
arrangement. 

EXAMPLES. 
XXXIII. Screw. Articles 236-242. 

1. "What weight can be raised by a power of 30 lbs. acting on a 
iever-arm 2 feet long, if the screw has 2 threads to the inch? 

2. If a power of 40 lbs. acting on an arm 25 inches long can 
support a weight of 8000 lbs., what will be the distance between 
the threads of the screw? 

3. A screw has 10 threads to the inch; the circumference 
described by the power is 4 feet : What power is needed to sup- 
port a weight of 6000 lbs? 

4. While the point of application of the power makes a revo- 
lution of 3 feet, the screw advances % of an inch; the power is 50 
lbs. : What is the weight raised? 

5. The angle of the screw is 10°, and the length of the power- 
arm is twenty times the radius of the cylinder : What is the me- 
chanical advantage? 

6. The circumference described by the power-arm is 20 feet, and 
the mechanical advantage 480 : How many threads in the screw 
are there to the inch ? 

7. The circumference described by the power-arm is 14 feet, the 
power is 60 lbs., and the weight 6 tons (6 X 2240 lbs.): What is 
the distance between the threads of the screw? 

8. The power-arm of a differential screw is 18 inches; there are 
6 threads to the inch in the larger screw, and 8 threads in the 
smaller; the power is 30 lbs. : What weight can be supported? 



CHAPTER IX.— PENDULUM. 

244. Motion in a Vertical Circle. Let ABC (Fig. 188) 
represent a vertical circle, regarded as perfectly smooth. 
Suppose a particle to start from rest at A and slide down 



*9' 




toward B ; its velocity (v) at any point M will be the 
same (40) as if it had fallen through the vertical height 
EF; that is, 

v' = 2g.EF (1) 

Since DMB and DAB are right angles, by geometry, 
MB* = DB.FB, (2) 

AB* = DB.EB. (3) 

Let AB = a, and MB = z; also, Z>5 = 2r; substitut- 
ing these values and subtracting (2) from (3), 

a? - z* = 2r (EB - FB) = 2r.EF. (4) 



245.] SIMPLE PENDULUM. 253 

Introducing the yalue of i^Ffrom (4) in (1), we obtain 

v * = l(a'-z>), v= ±\/l(a'-z'). (5) 

For the point A (or A f ), z = a, and therefore v = 0; 
for B, z = and v = ± a y — , the double sign indicat- 



t „ fh pf +Ti p TYi n+,i nn m a.v hp f rri 



ing that the motion may be from A toward A' (-f), or the 
reverse (— ). For two points M and if', equally dis- 
tant from B, BM = + z, and BM ' — — z\ for both 
these the yalue of v is the same, and for each there is a 
-f- value and a — value, according to the direction of 
the motion. It is evident that if the particle were pro- 



jected from B with a velocity equal to a \ — , it would 
ascend to A' before coming to rest. 

245. Motion of a Simple Pendulum. An ideal simple 
pendulum is a material particle attached by a string 
without weight to the point of suspension, and vibrating 
without resistance from friction or any other source. A 
particle so suspended will, if set in motion, continue to 
vibrate to and fro in an arc of a circle, and will follow 
the same laws as the body descending the smooth curve 
considered in the previous article. The tension of the 
string, like the resistance of the plane, is always equal 
to one component of the weight, and is in both cases 
exerted at right angles to the direction of motion, and 
hence does not influence the velocity of the particle. 

Therefore the value of the velocity, v — y — (a? — z*) 9 
obtained in Art. 244 will apply also to the pendulum at 



254 



PENDULUM. 



[245. 



any point in its course. If now the radius r (the length 
of the pendulum) is great and the length of the arc of 
vibration is very small, we may take a and z in this 
value of v as representing the arcs instead of the chords, 
and this is assumed in the following demonstration. 
The error arising from this assumption may be neglected 
without destroying the value of the result. 

Let aa' (Fig. 189) represent the arc of vibration of the 
simple pendulum, whose length (r) is CB. Take the 
j, straight line AA' equal to aa' , 

and so that every point m on the 
arc has a corresponding point M 
on the straight line. We may 
without error imagine the pen- 
dulum to vibrate from A to A', 
following the same law as in the 
arc aa', so that its velocity at 
any point M will be expressed by 
the above formula, 

when AB = BA' = a, and BM 
= z. Here, as remarked above, 
a and z are arcs, not chords. 

Upon A A' as a diameter de- 
scribe the circle ANA' , and sup- 
pose a particle to move uniformly 
A about the semi-circumference 
ANA' with the constant velocity 




Then, since the dis- 



Fig. 189. 



tance passed over is na, the time 
(t) required for this imaginary particle to go from A to 



245.] SIMPLE PENDULUM. 255 

A', since the distance and Telocity are known, is (19) as 
follows : 

, s na A /r 

t = — = - — — — n y — . 



9 9 



a y\ 



It will now be shown that this expression also gives the 
time required by the pendulum to vibrate from A to A' 
(that is, from a to a'). 
The constant velocity V of the imaginary particle, viz. 

ay—, may be represented (20) at any point in its path, 

as N, by a tangent to the curve NQ l=ay—). The 

horizontal component of this velocity is then represented 
by NP. 

NP = NQ. cos QNP, 



= ay ^-.cos BNM. 
r 

But a cos BNM = NM, and 

NM=z ^NB* - BM 2 = Va 2 - z\ 

Therefore NP = \. g - (a 2 - z 2 ). 

Now, this expression for the horizontal component of 
the velocity of the imaginary particle at N is also 
the value of the velocity of the pendulum at the cor- 
responding point M. The meaning of this result is 
as follows: If, at the same instant that the pendulum 



256 PENDULUM. |245- 

commences to vibrate from A to A', the imaginary 
particle starts from A about the semi-circumference 
with its constant Telocity, the velocity of the pen- 
dulum at any point (as M ) will be the same as the 
horizontal component of the particle at the correspond- 
ing point (as N), and therefore the two will reach A f 
at the same time. The time required for a single vibra- 
tion of the pendulum is, therefore, the same as that for 
the imaginary particle to go in its path from A to A'; but 



the latter has been shown to be equal to n Y — , hence 

this expression also gives the time of the pendulum. If 
for r we write I, the length of the pendulum, we obtain 

9 
It is to be noted that: 

(1) The time of a vibration is independent of the 
length of the arc, when it is taken very small. That is, 
the pendulum will vibrate through an arc of 1, 2, or 
3 degrees in sensibly the same time. Further, when the 
arc of vibration is constant (as in the clock) the times 
of vibration are necessarily the same. 

The motion of the pendulum, to and fro in a small 
arc, is the type of a great variety of isochronous vibra- 
tions; for example, those of a tuning fork, of a musical 
string, of the particles of air in sound-waves, and, too, of 
the ether (109) particles in the case of heat and light. 

(2) If g is constant— that is, for the same point on the 
earth — the time of vibration varies directly as the square 
root of the length (/a^orfal). If a given pendulum 
vibrates in 1 second, one four times as long will vibrate 
in 2 seconds, and another one quarter as long will 
vibrate \ a second. 



246.] COMPOUND PENDULUM. 257 

(3) If the length is constant, the time of vibration 
varies inversely as the square root of the force of grav- 

(4) If the time is constant — that is, for two pendu- 
lums at different stations, each vibrating in 1 second — 
the length varies directly as the force of gravity (I oc g). 

The relations in (3) and (4) give a ready method of 
comparing the values of g for different latitudes or for 
different heights above the sea-level, as further explained 
in Art. 248. 

246. Compound Pendulum. The simple pendulum, 
used in the preceding demonstration, is an ideal form 
not obtainable for actual experiment. Any pendulum 
that can be constructed is a compound pendulum, con- 
sisting of an indefinite number of material particles 
rigidly joined together and vibrating from a fixed axis. 
In order to apply the result in Art. 245 to a compound 
pendulum, it is necessary to obtain the value of I, and 
for this end use is made of the principle: 

The length of a compound, pendulum is equal to the 
length of a simple pendulum which would vibrate in the 
same time. 

Consider the compound pendulum as a simple metal 
rod: it is obvious that the particles near the axis of 
support, called the axis of suspension, tend to vibrate 
more quickly, and those farthest from it more slowly, 
than they can do, since they must all vibrate in the 
same time. It is obvious that there must exist a line in 
the rod which is so situated that the motion of the 
particles on it is neither quickened nor retarded by the 
rest ; that is, these particles would, if alone, vibrate in 
the same time as that of the bar as a whole. This line 



<y & 



258 PENDULUM. [247. 

through these particles and parallel to the axis of sus- 
pension is called the axis of oscillation. The distance 
between these two axes is the length of the compound 
pendulum as above defined. 

247. To find the axis of oscillation, use is made of 
the principle : The axis of suspension and axis of oscilla- 
tion are interchangeable. Therefore, if the pendulum be 
swung on one axis and the time of vibration be deter- 
mined, and then the second axis be found, so that 
' ] if this is made the axis of suspension the time of 
I "vibration will be exactly the same, the latter axis 
is the axis of oscillation. The truth of this prin- 
ciple is established in works on higher Mechanics. 
The pendulum exhibited in Fig. 190 is a form 
devised by Kater : a and T> are the two axes of 
X suspension and oscillation; v and w are two 
slides, the position of which may be adjusted 
until the condition in regard to the equal times of 
vibration for the two axes is satisfied. 

248. Application of the Pendulum. The most 
important application of the pendulum is as an 
instrument for determining the value of the 
acceleration of gravity (g). The direct determi- 
nation of g requires the observation of either 
the velocity acquired (32 feet per second in a 
second) or the space passed over (16 feet in the 
first second from rest) by a body (27) falling in a 
vacuum. This method is obviously impractica- 
Fiaiso. We; If Attwood > s machine (74) be employed, 
the force of gravity may, as it were, be weakened so that 
the velocity acquired and space passed through are much 
less than for a body falling freely. In this way approxi- 
mate values of g may be obtained. 



250.] APPLICATION OF THE PENDULUM. 259 

For accuracy, however, the seconds pendulum gives 
the simplest and most satisfactory method of determin- 
ing the value of g. From Art. 245 we see that 

A'fi nH 

t=7ty- 9 or 9 = ~f- 

For the seconds pendulum, g = n:*l. 

In the actual determination of the value of I for a 
pendulum vibrating in seconds on any point of the 
earth's surface, many refinements of observation are 
required. It is sufficient, however, to say here that 
when the experiments are carried on with all possible 
care, and when the many necessary corrections have been 
introduced, a very high degree of accuracy is obtainable 
for the value of I, and consequently of g. 

249. The following table gives the lengths of the 
seconds pendulum and the value of g (for the sea-level) 
for some important points on the earth. Those for the 
equator and pole are calculated from an equation the 
constants of which have been deduced from numerous 
pendulum experiments in different localities; the others 
are from direct observations (U. S. Coast Survey) : 

Value of g, v* w nf J 

Latitude. in feet-per-sec- £?£?£"' 

ond per second. mincnes. 

Equator 0° 32.091 39.017 

New York (Hoboken) 40° 43' 32.161 39.103 

Paris 48° 50' 32.185 39.133 

London (Kew) 51° 29' 32.193 39.142 

Berlin 52° 30' 32.195 39.144 

Pole 90° 0' 32.255 39.217 

250. The lengths of the seconds pendulum, as deter- 
mined by numerous experiments, are of great value, as 
giving the means of comparing the intensity of the force 



260 PENDULUM. [251. 

of gravity at the different stations. By this means the 
ellipticity of the earth has been determined. A formula, 
alluded to in the previous article, has also been obtained 
which gives the values of I and g for any required lati- 
tude. The ellipticity of the earth, obtained in this 
way, differs somewhat from the similar result from the 
trigonometrical measurements of long arcs of meridians. 
Moreover, the values of I and g derived from the formula 
do not always agree with those obtained by direct ex- 
periment as closely as might be expected. The ex- 
planation of the variations is to be found in the facts 
(1) that the earth is rather an ellipsoid than a spheroid, 
and, further (2), that the density of the earth's crust at 
different points is not uniform. This latter point has 
been extensively investigated by means of these pendu- 
lum observations; it is found, for example, that the attrac- 
tion on the coast is greater than in the interior, and 
that it is still greater on islands in the sea. From this 
it is argued that the density of the earth's crust under 
the ocean is greater than the average of that forming 
the dry land. Other similar results have also been 
obtained by this means. 

251. It has also been proposed to make the pendulum 
a basis of a system of weights and measures, on the 
ground that it would, at a given place, be a standard 
which could be at any time replaced if others were 
destroyed. It was, for example, enacted by Parliament 
in 1824 that the length of the standard yard should bear 
to the length of the seconds pendulum in London (in 
vacuum and at the sea-level) the ratio of 36 : 39.1393. 
The difficulty in obtaining the length of the pendulum 
with the degree of accuracy now needed for a standard 
of measures makes the relation of little practical value 



251.] EXAMPLES. 261 

The pendulum finds a further application as a regu- 
lator for clocks. 

EXAMPLES. 

XXXIV. Pendulum. Articles 244-247. 

[The length of the seconds pendulum at New York is about 39.10 
inches, and g = 32.16.] 

1. Required the length of a pendulum at New York which 
will vibrate (a) in £ second, (&) in 2 seconds, (c) in %\ seconds. 

2. Required the length of the seconds pendulum where the 
acceleration of gravity is 32.25 (that is, near the pole). 

3. Required the length of a pendulum to vibrate in 2 seconds, 
where the value of g is 32.1. 

4. What would be the length of a pendulum to vibrate in 
1 second at the surface of the sun (acceleration of gravity = 28 #)? 

5. How many beats in a minute would a pendulum 8 feet long 
make in New York? 

6. If a pendulum 24 inches long vibrates in £ second, what is 
the length of a seconds pendulum? 



ADDITIONAL EXAMPLES, 

INTRODUCING THE METRIC UNITS. 



I. Uniform Motion of Translation or Rotation. Articles 17-21; 

pp. 9-12. 

1. A body travels 10 meters per second : How far will it go in 
a day of 24 hours? 

2. A velocity of 50 kilometers per hour corresponds to a rate 
of how many meters per second? 

3. A man walks uniformly 6 kilometers per hour: (a) How 
many decimeters does he go in a second? (b) How many meters 
in a minute? 

4 Two bodies start from the same point in opposite directions; 
the one moves at a rate of 3 meters per second, the other at a 
rate of 30 kilometers per hour: (a) What will be the distance 
between them at the end of 8 minutes? (b) When will they be 
6 kilometers apart? 

5. How far will the bodies in the preceding example be apart 
at the end of the same time, if they move in the same direction? 

6. (a) What is the angular velocity of a revolving wheel having 
a radius of 1 meter, if it makes 14 revolutions per second (take 
jr = 8f)? (b) How far (in kilometers) will a point on the circum- 
ference travel in 10 hours? 

II. Uniformly Accelerated Motion Articles 22-28 ; pp. 13-20. 
A. Falling Bodies (g — about 9.80 meters at New York). 

[The body is supposed to start from rest.] 
1. Calculate the distances fallen through (in meters) and the 
acquired velocities in 1, 2, 3, 4 seconds from rest. 



264 

2. A body falls 10 seconds : Required (a) the velocity acquired ; 
(b) the whole distance fallen through; (c) the space passed over in 
the last second of its fall. 

3. A body has fallen through 90 meters: Required (a) the time 
of falling; (b) the final velocity. 

4. A body has acquired in falling a velocity of 73.5 meters per 
second: Required (a) the time of falling; (b) the distance fallen 
through. 

5. A body in falling passed over 44. 1 meters in the last second : 
Required (a) the time of falling; (b) the distance fallen. 

B. General Case. — Acceleration = f. 

1. A body moves 100 meters in the first 5 seconds from rest: 
What is the acceleration? 

2. A body moves 10 meters in the first second: (a) What is the 
acceleration? (b) How far will it go in 6 seconds? (c) What will 
be its final velocity at the end of this time? 

3. The acceleration is 12 meters-per-second per second: (a) 
What velocity does a body acquire in 5 seconds? (b) What space 
does it pass over? 

4. A body moves 54 meters in 3 seconds, and 96 in the next 2: 
Is its motion uniformly accelerated? 

5. A body passes over 64 meters in 4 seconds: What distance 
must it go in the next 5 to satisfy the condition of uniformly 
accelerated motion? 

6. The velocity of a body is increased from 20 to 40 meters per 
second while it passes over 30 meters: What is the acceleration? 

III. Composition of Constant Velocities. Articles 29-37; pp. 22-27. 

1. The velocity of a steamboat is 10 kilometers per hour, that 
of the stream is 8, and a man walks the deck from stern to 
bow at the rate of 6 : Required the actual velocity of the boat 
(a) if headed up stream, and (b) down stream; also (c, d), that of 
the man in each case. 

2. The velocities of the boat and stream are respectively 100 
meters and 80 meters per minute, and the boat is headed directly 
across the stream (Fig. 11): (a) What will be the actual direction 
of the boat's motion? (b) What the rate of its motion? (c) How 
long will the passage take if the stream is 2 kilometers in width? 



INTRODUCING THE METRIC UNITS. 265 

3. The velocities of boat and stream are as in 2, but it is re- 
quired that the boat shall go directly across from A to G (Fig. 
12): {a) In what direction must the boat be headed? (b) What 
will be its actual velocity across? (c) What time will the passage 
take, the width being as in 2? 

4. A bull on a horizontal surface tends to move north with a 
velocity of 12 meters per second, and east with a velocity of 
5 meters per second: (a) What will be the actual velocity, and (b) 
in what direction? 

5. A ball, moving north at a rate of 8 meters per second, re- 
ceives an impulse tending to make it move due south-east with 
the same velocity: (a) What path will it take, and (b) at what 
rate will it move? 

6. A man, skating uniformly at a rate of 4 meters per second, 
projects a ball on the ice in a direction at right angles to his 
motion at a rate of 3 meters per second: What is (a) the actual 
rate, and {b) the direction of its motion (friction neglected)? 

IV. Resolution of Constant Velocities. Article 38 ; pp. 28, 29. 

1. A ball tends to move in a certain direction at a rate of 6 
meters per second, but it is constrained to move at an angle of 
60° with this direction: Required its velocity in the latter direc- 
tion. 

2. A ball rolls at the rate of 6 meters per second across the 
diagonal of a rectangular room ABGD whose dimensions are 
15 X 20 (= AB X AG): What is its rate of motion parallel to 
each side? 

3. A body moves N. 30° E. at a rate of 5 kilometers per hour: 
Required its rate of motion northerly and easterly. 

4. A boat, though headed directly across a stream, actually 
moves diagonally across the stream at an angle of 30° (BAG, Fig. 
11) and at a rate of 15 kilometers per hour: Required (a) the rate 
of the boat, and (b) of the current taken independently. 

V. Falling dozen an Inclined Plane. Article 40; pp. 32, 33. 

[The plane is supposed to be perfectly smooth, so that there is no 
friction.] 
1. The angle of the plane is 30° : Required (a) the acceleration 



266 ADDITIONAL EXAMPLES, 

down the plane; (b) the distance fallen through in 4 seconds; (c) 
the velocity acquired. 

2. The height of the plane is 19.6 meters and the length 78.4: 

(a) What is the time required to reach the bottom? (b) What is 
the velocity acquired? 

3. The angle of the plane is 45° : Required the time of falling 
490 meters. 

4. The length of a plane is 630 meters ; a body falls down it in 
30 seconds: (a) What is the acceleration? (b) What is the height 
of the plane? 

VI. Bodies projected vertically downward. Articles 41, 42; 
pp. 34, 35. 

1. A body is thrown vertically down with an initial velocity of 
12 meters per second : Required (a) the velocity at the end of 7 
seconds ; (b) the distance fallen through. 

2. A body is projected down with an initial velocity of 19.1 
meters per second: (a) How long will it require to fall 218 meters? 

(b) What velocity will it then have ? 

3. What velocity of projection must a stone have to reach the 
bottom of a cliff 100 meters high in 3 seconds? 

4. A stone is dropped from a bucket which is descending a 
shaft at the uniform rate of 3.5 meters per second, and at the 
moment when the bucket is 75 meters from the bottom : (a) How 
far will they be apart in two seconds? (b) When will the stone 
reach the bottom? 

VII. Bodies projected vertically upward. Article 44; pp. 37-39. 

1. The velocity of the projection upward is 49 meters per 
second: Required (a) the time of ascent; (b) of descent; (c) the 
height of ascent; (d) the distance gone in the first and last 
seconds of ascent. 

2. A body is projected up with a velocity of 42 meters per 
second: (a) When will it be 37.1 meters above the starting-point? 

3. What velocity of projection must a ball have in order to 
ascend just 160 meters? 

4. What time does a body require to ascend 250 meters, that 
being the highest point reached? 



INTRODUCING THE METRIC UNITS. 267 

5. A ball thrown up passes a staging 36.4 meters from the 
ground at the end of 2 seconds: (a) What was the velocity of the 
projection? (b) When will it pass it again? 

VIII. Projected up or down a smooth Inclined Plane. Article 45 ; 

p. 39. 

1. The height of the plane is 105 meters, the length is 420 
meters, the velocity of projection down is 20.3 meters per second: 

(a) How long will it require to descend? (b) What will be the 
final velocity? 

2. The angle of the plane is 30°, the velocity of projection 
down is 10 meters per second: Required (a) the velocity at the 
end of 4 seconds: (b) the distance gone through. 

3. The height and length of the plane are 160 and 320 meters 
respectively : (a) What velocity is required that the body should 
just reach the top? (b) What time is needed? 

IX. Bodies projected against Friction. Articles, 41, 42; pp. 34, 35. 

[The retardation (or minus acceleration) due to friction takes the 
place of the fin the formulas of articles 42 and 44.] 

1. A body projected on a rough horizontal plane has at starting 
a velocity of 40 meters per second, but loses this at the rate of 4 
meters for each succeeding second : (a) What is the retardation 
(minus acceleration) due to friction? (b) When will the body 
stop? (c) How far will it have gone? 

2. The retardation due to friction is for each second 6 deci- 
meters per second for a given sliding body, the initial velocity 
is 12 meters per second: Required («)the time it will continue to 
slide ; (b) the distance it will go ; (c) its velocity at the end of 3 
seconds. 

3. If the retardation of friction is 2 decimeters-per-second per 
second: (a) What initial velocity (in kilometers per hour) must a 
body have in order to slide just 160 meters? (b) If the velocity is 
doubled, how much farther will it go? 

X. Projectiles. Articles 47-51 ; pp. 43-50. 

1. The initial velocity of a projectile is 245 meters per second, 
and the angle of elevation is 30° : Required (a) the time of flight; 

(b) the range. 



268 



2. The initial velocity is 140 meters per second : What angle of 
elevation will give a range of 1.5 kilometers? Show that there 
are two answers. 

3. The angle of elevation is 15° : What initial velocity is re- 
quired that the range should be 4 kilometers? 

4 A rifle-ball is shot horizontally from the top of a tower 44.1 
meters high, and with an initial velocity of 400 meters per 
second : When and how far from the base of the tower will it 
strike the horizontal plane below? 

5. A ball is thrown horizontally from the top of a cliff above 
the sea; it strikes the w r ater in 5 seconds, and at a horizontal dis- 
tance of a kilometer: What was (a) the initial velocity, and (b) 
what was the height of the cliff? 

XI. Mass — Density — Volume. Article 56 ; pp. 53, 54. 

1. What is the ratio in volume of a piece of silver weighing 
20 kilograms and having a density of 10.5 (referred to water as 
unity), and a piece of iron weighing 5 kilograms and having 
a density of 7 ? 

2. What is the ratio in weight (that is, in mass) of two blocks 
of stone, one having a volume of 1 cubic decimeter and a density 
of 3, the other a volume of 400 cubic centimeters and a density 
of 2.75 ? 

3. If a liter of dry air at 0° weighs 1.2932 grams, and a liter of 
water at the same temperature weighs 999.88 grams, what is the 
density of the water referred to that of air as unity ? 

4. What is the weight of a liter of mercury at 100°, the expan- 
sion of volume from 0° to 100° being in the ratio of 1 : 1.0154 ? 
The density of mercury at 0° is 13.6, referred to water as unity. 

XII. Force of Gravity. Articles 63-65; pp. 58-62. 

1. At what distance from the centre of the earth would a mass 
of matter weighing 16 kilograms on the earth's surface exert a 
full equivalent to 1 kilogram on a spring-balance ? 

2. If the mass of the sun is 355,000 times that of the earth, and 
its diameter 112 times, what would be the acceleration of gravity 
at its surface ? 

3 If the moon's mass is -^ of that of the earth, and its dia- 



INTKODtTCmG THE METEIC UNITS. 269 

meter 3476 kilometers, that of the earth being about 12,715 kilo- 
meters, what is the acceleration of gravity on the moon's surface? 
4. If the distance from the earth to the moon is 60 times the 
earth's radius, what is the force of the earth's attraction at the 
moon ? 

XIII. Collision of Inelastic Bodies. Article 70; pp. 70, 71. 

[The bodies are supposed to be perfectly inelastic, their motion 
is uniform, and the impact is direct.] 

1. A ball weighing 10 kilos and having a velocity of 6 meters 
per second overtakes a second ball weighing 5 kilos and whose 
velocity is 3 meters per second : What is the final velocity ? 

2. If the first ball in the preceding example meets the second, 
what is the final velocity? 

3. A body weighing 40 kilos strikes another at rest weighing 
360 kilos, and the two move on with a velocity of 1 meter per 
second: What was the original velocity of the first ball? 

4. Three bodies, each weighing 4 kilos, are situated in a 
straight line ; a fourth, weighing 8 kilos and moving at a rate of 
6 meters per second, strikes them in succession: What velocity 
results after each impact? 

5. A rifle-bullet weighing 30 grams is fired into a suspended 
block weighing 15 kilos; the blow causes the wood to rise 36 
millimeters : Required the velocity of the bullet at the moment 
of impact. 

XIV. General Dynamical Pivblems. Articles 68, 73-76; pp. 66, 
67, and 75-80. 

1. If in Attwood's machine P= 102 grams and Q = 45 grams: 
(a) What is the acceleration? (b) What space will be passed 
through in 2 seconds ? 

2. (a) At what height above the earth's surface would a body 
fall 625 millimeters in the first second from rest? (b) If its weight 
was 16 kilos, what pull would it exert on a spring-balance at this 
point? 

3. A weight of 6 kilos hanging over the edge of a smooth table 
drags a weight of 15 kilos with it : What is the acceleration and 
the tension of the string? 



270 ADDITIONAL EXAMPLES, 

4. For what time must a force of 60 grams (gravitation meas- 
ure) act on a body weighing 2 kilos to give it a velocity of 6 
metres per second ? 

5. A body weighing 140 kilos is moved by a constant force, 
which generates a velocity of 2 meters per second in one second : 
What weight could the force support? 

6. What constant force (a) in gravitation measure (grams), (b) 
in absolute measure (dynes), will cause a body weighing 490 
grams to pass over 400 meters in 10 seconds on a smooth hori- 
zontal surface? 

7. A constant force of 980 dynes gives a body an acceleration 
of 7 meters per second in one second: What is the weight of the 
body (in grams)? 

8. What weight could a force equal to 1 dyne support? 

XV. Centripetal and Centrifugal Forces. Articles 77-81 ; pp. 83-8. 

1. A ball weighing 10 kilos is whirled by means of a string 
around a centre at a radius of 2 meters, with a linear velocity of 
7 meters per second : What is the value of /, and what is the 
tension of the string (F)1 

2. A ball weighing 14 kilos attached to a centre at a distance 
of 3 meters makes 420 revolutions in a minute {it = 3|) : What is 
the pull on the centre? 

XVI. Friction. Articles 82-94; pp. 90-98. 

1. A force of 6 kilos is just sufficient to move a body weighing 
48 kilos uniformly along a horizontal plane: What is the coeffi- 
cient of friction? 

2. The value of ju is .3, the weight of the body is 16 kilos: 
What force is required to move it uniformly? 

3. It is found that a force of 40 grams suffices to move a body 
uniformly on a horizontal surface, where the value of the co- 
efficient of friction is known to be .25: What is the weight of the 
body? 

4. A body weighing 15 kilos is just on the point of sliding 
when the surface it rests upon is inclined 20° : (a) What is the co- 
efficient of friction and the force of friction? (b) If the weight of 
the body is doubled, what values have these quantities? 



INTRODUCING THE METRIC UNITS. 271 

5. A body weighing 12 kilos rests on an inclined plane whose 
angle of inclination is 30° and where /x = .6: What is the force 
of friction? 

XVII. Work. Articles 95-100; pp. 101-105. 

[The unit of work is usually taken as one kilogram-meter; on 
the C. G\ S. system it is an erg, or one dyne-centimeter.'] 

1. How many foot-pounds correspond to one kilogram-meter? 

2. A weight of 300 kilos is raised to the top of an inclined 
plane whose length is 1200 metres, and the angle of inclina- 
tion = 10°: What work is done? 

3. A sled weighing 600 kilos is dragged 15 kilometers on the 
snow, where the coefficient of friction is .075: What work is 
done against friction ? 

4. How much work is done against friction in dragging a 
weight of 200 kilos a distance of 1000 meters along a horizontal 
plane, if the coefficient of friction is .5? 

5. A weight of 100 kilos is dragged up an inclined plane whose 
length is 2600 meters, and whose height is 1000 meters (jx = .3): 
How much work is done ? 

XVIII. Potential and Kinetic Energy. Articles 101-118; 
pp. 106-124. 

1. How many kilogram-meters of work are equivalent to one 
heat-unit {i.e., to raise 1 kilo of water 1° C.)? 

2. The weights of a clock weigh 20 kilos, and they have 10 
meters to fall* How much work do they represent when wound 
up ? 

3. A mill-pond has a surface of 1000 square meters and an 
average depth of 1 meter; supposing it 50 meters above the sea- 
level, how much potential energy does it represent? 

4. How much work is accumulated or stored up (= kinetic 
energy) in a cannon-ball weighing 100 kilos and moving at a rate 
of 280 meters per second? How much heat will be generated if 
its mass motion is entirely destroyed by the impact with the 
target? 

5. A bullet weighing 30 grams has a velocity of 420 meters per 
second: How much woik can it do? 



272 



6. A body weighing 12 kilos is projected along a rough hori- 
zontal plane (// = .25) with an initial velocity of 140 meters 
per second: How far {a) will it go before coming to rest, and 
how long (b) will it slide? 

7. A hammer weighing 5 kilos and moving with a velocity of 
1.4 meters per second drives a nail into a plank 1 centimeter: 
What resistance does it overcome? 

8. A weight of 500 kilos, used as a pile-driver, falls 6 meters 
and drives the pile in 2 centimeters: What resistance does it 
overcome? 

XIX. Parallelogram of Forces. Articles 124-136; pp. 129-142. 

1. Two forces, P = 70 grams, Q = 240 grams, act at right 
angles to each other : Required the magnitude and the direction 
of their resultant. 

2. Two forces, P = 12 kilos, Q = 10 kilos, act at an angle of 
120° : Required the magnitude of R. 

3. Of two forces, P = 12 and Q = 24 kilos, the angle between 
Q and R is 30° : Required R and the angle between P and Q. 

4. A peg in a wall is pulled by two strings with forces of 4 
kilos each; they are equally inclined downward (40°) to the 
vertical: What weight hung on the peg would give an equal 
strain? 

5. A peg in a wall is pulled by two strings, one horizontal with 
a tension of 350 grams, and the other vertical with a tension of 
840 grams: What single force would exert an equal pull upon it? 

6. A weight is supported by two equal strings attached to nails 
in the ceiling and enclosing an angle of 120°; the tension of each 
string is 8 kilos: What is the weight supported? 

XX. Resolution of Forces. Articles 137, 138; pp. 143-146. 

1. A force of 60 kilos is exerted in a direction N. 20° E. : What 
portion of it is felt north? what portion east? 

2. A weight of 12 kilos is supported by two strings at an angle 
of 120° ; one (a) goes (Fig. 66, p. 144) horizontally to the vertical 
wall, and the other (b) to the ceiling: What is the tension of the 
two strings? 

3. A picture, whose weight is 40 kilos, is supported by a cord 



INTRODUCING THE METRIC UNITS. 273 

attached to the upper corners and carried over a nail so as to 
include an angle of 75°. If the top of the picture is horizontal, 
what are the tensions of the strings? 

4. A horse drags a sled by a rope inclined at the ground at an 
angle of 10° ; the tension of the rope is 300 kilos : What is the 
effective component of the force exerted ? 

5. A weight of 640 grams is supported by two strings, one of 
which makes an angle of 30° with the vertical, and the other 60° : 
Find the tension of each string. 

XXI. Resolution of Forces along two Rectangular Axes. 
Articles 140, 141, 147-149; pp. 147-149, 156-158. 

1. Find the magnitude and direction of the resultant of the 
following forces: P = 100 kilos, Q = 50, 8 = 200. The angle be- 
tween P and Q = 60°, between Q and 8 = 270°. 

2. Required the magnitude and direction of the resultant of the 
following forces: P — Q = 8 = T — 100 kilos. The angles are 
as follows: between P and Q = 60°, between Q and 8 = 120°, 
between 8 and T = 120°. 

XXII. Parallel Forces. Articles 143-149; pp. 152-159. 

1. A rigid rod, supported at the ends A and B, has a weight of 
24 kilos hung 2 meters from A and 4 meters from B: What pres- 
sures do the supports feel? The weight of the rod itself is neg- 
lected here, as, too, in the following examples. 

2. ABC is a rigid rod; at B a weight Wis hung, so that AB — 
24 and BG = 32; the pressure at A is 14 kilos: What is the pres- 
sure at C, and what is W ? 

3. A weight of 80 kilos is carried by means of a rigid rod on 
the shoulders (at the same height) of two men A and B; the dis- 
tances from them are 2 and 3 meters respectively: What weight 
does each carry? 

4. A table has as its top an equilateral triangle ABG (Fig. 82); 
a weight of 10 kilos is placed at O, so that the perpendicular dis- 
tance from O on BC = 18 centimeters, and those on AC, AB 
each equal 36 centimeters: What is the pressure on each of the 
three legs? 

5 A rod 40 centimeters long and whose weight acts at its 



274 

middle point rests on two vertical props placed at the ends: 
Where must a weight, equal to twice that of the rod, be placed 
that the pressure on the props shall be as 4 : 1 ? 

XXIII. Moments. Articles 151-156; pp. 161-167. 

1. A force, P = 24 kilos, acts at right angles to an arm 3 meters 
long: What is its moment? 

2. A rigid rod AB, 2 meters long and free to turn about B, is 
acted on by a force, P = 30 kilos, whose direction makes an angle 
of 30° with AB: What is the moment of P ? If the angle is 150°, 
what is the moment? 

3. A force, P = 100 kilos, acts at the extremity of a rod, AB, 
4 meters long, and at an angle of 120° : What is the moment of P 
about B ? 

4. A bar 5 meters long and pivoted at the middle has a weight 
of 10 kilos hung at one extremity: What is the moment of the 
weight (a) when the bar is horizontal, (b) when it makes an angle 
of 20° below, and (c) of 70° above with the horizontal position? 

XXIV., XXV. Centre of Gravity— Stability. Articles 159-177; 
pp. 169-189. 

1. Where is the centre of gravity of two bodies, A and B, 
weighing 30 and 42 grams respectively, rigidly connected by a 
weightless rod 36 centimeters long? 

2. A rod AB, 1 meter long and weighing 250 grams, has a 
weight P = 4£ kilos hung at the end B : Where will it balance? 

3. What weight must be hung at the end of a rod \ meter long 
and weighing 30 grams that it may balance 25 millimeters from 
that end? 

4. A rod 2 meters long and having a weight of 5 kilos at one 
end balances at a point 2 millimeters from this end: What is its 
weight? 

5. A uniform rod AB, 1 meter long and weighing 1 kilo, has a 
weight of 750 grams at the end A, and one of 250 at the end B: 
Where will it balance? 

6. A uniform metal wire is bent into the form of an isosceles 
triangle, ABC, so that AB= AC — 117 millimeters, and BC= 90 
millimeters: Where is the centre of gravity? 



INTKODUCING THE METEIC UNITS. 275 

7. A table 2 meters square stands upon four legs, each of which 
is 300 millimeters in from the adjacent edges; its height is 800 
millimeters and its weight 24 kilos: "What is the least force 
required to put it on the point of overturning if applied at the 
edge (a) as a horizontal push? (b) as a pressure directly down? 

8. A table, having a circular top of £ meter radius, is sup- 
ported on four legs placed at the edge and at equal distances 
from one another; the height is £ meter and the weight 16 kilos: 
"What is the least force that will put it on the point of turning it 
if applied at the top (a) as a horizontal push? (b) as a pressure 
down? (c) acting vertically upward? 

9. "What work would be done in overturning a cylindrical 
column of stone weighing 10,000 kilos, 3 meters high and 1 meter 
in diameter, supposing that the centre of gravity is on the axis 
(a) at the middle? (b) .25 meter from bottom, (c) the same distance 
from top? 

XXVI. Lever. Articles 185-190; pp. 195-201. 

[The weight of the lever is to be neglected, except when other- 
wise stated.] 

1. The force P= 40 kilos acts as in Fig. 126, p. 196; AF = 
2 meters and AB = 3£ meters: "What weight can be supported? 

2. If (Fig. 127, p. 196) AB=5,BF=2i meters, and the weight 
is 60 kilos, what force Pis required to support it? 

3. If (Fig. 128, p. 196) AB= 14, AF= 2 meters, and P = 24 
kilos, what weight can P support? 

4. AFC is a bent lever (Fig. 129, p. 197); AF= 12, FC=-U, 
AFC=n§°, P=20 kilos: What is TF? 

5. CFD is a bent lever (Fig. 131, p. 197); CF= 16, Pi) = 24, 
PDF= 150°, FGW= 120°, and W= 60 kilos: "What is P ? 

6. A heavy uniform rod DF (Fig. 133, p. 199), weighing 10 
Mios and f meter long, is hinged at F ; it is supported by a 
string carried from D to a point E, 250 milimeters vertically 
above F: "What is the tension of the string? 

7. A heavy uniform shelf DF (Fig. 134), \ meter wide, weigh- 
ing 24 kilos, and hinged at F, is supported by a prop carried 
from C (CF= 400 millimeters) to a point E below F, so that 
CF= FE: What pressure does this prop feel? 



276 

8. Forces of 8 and 12 kilos act at the extremities of a straight 
bar 4 meters long, and in directions making angles of 120° and 
150° respectively with it: Where is the fulcrum in case of equi- 
librium? 

XXVII., XXVIII. Balance— Steelyard. Articles 191-196; 
pp. 202-210. 

1. A body is equivalent to a weight of 6 kilos in one pan of a 
false balance, and of 6£ kilos in the other: What is the true 
weight? 

2. A body is equivalent to a weight of 84 grams from one arm 
of a false balance, and of 72 grams from the other : What is the 
ratio of the lengths of the arms? 

3. The true weight of a body is 1 gram, its apparent weight in 
one pan of a balance is 960 milligrams: What would it seem to 
weigh in the other pan? 

4. The whole length of a steelyard is f meter: CO (Fig. 138) = 6 
millimeters, GA = 20 millimeters, P= 250 grams, and Q = i kilo: 
(a) Where is the zero of the scale? (b) What is the length of 
graduation for 1 kilo? (c) How large weights can it be used for? 

5. The weight of the beam of a steelyard is 1 kilo, and the 
distance of its centre of gravity is 16 millimeters from the ful- 
crum: Where must a counterpoise of 640 grams be placed to 
balance it? 

6. The length of a Danish steelyard is ^ meter, its weight is 
1£ kilos, and it acts at a point 50 millimeters from one end; a 
body weighing 6 kilos hangs at the other end: Where is the 
fulcrum? 

XXIX., XXX. Wheel and Axle — Pulley. Articles 202-226; 
pp. 216-234. 

1. The radius of the axle is 50 millimeters, that of the wheel 
is £ meter, and the power acting is 60 kilos: What weight is 
supported? 

2. A man, exerting a force of 40 kilos on a lever-arm 1J meters 
long, turns a capstan ; the radius of the circle about which the 
rope is wound is 150 millimeters: What is the pull felt upon the 
anchor? 



INTRODUCING THE METRIC UNITS. 277 

3. A weight of 300 kilos hangs by a rope 20 millimeters in 
thickness; r = .2 meter, and B = 1.25 meters; the power acts on 
a lever-arm without a rope: What is P ? 

4. A power of 40 kilos balances a weight of 600 kilos; the 
radius of the axle is 75 millimeters: What is the diameter of the 
wheel? 



5. In a combination of pulleys, as in Fig. 162, W= 704 kilos, 
and P = 4A kilos : How many pulleys are there ? 

6. In a combination of pulleys, as in Fig. 163, W= 288 kilos 
and P= 48 kilos: How many pulleys are there? 

7. In a combination as in Fig. 164, W = 837 kilos, P=27 
kilos: What is the number of pulleys? 

8. In the single movable pulley, W— 100 kilos: Calculate the 
value of P when 2a = 45°, = 135°. 

9. What force is needed to support 200 kilos by the second sys- 
tem of pulleys, there being 4 in all? What is the force if each 
pulley weighs -£- kilo ? 

XXXI. Inclined Plane. Articles 227-232 ; pp. 235-241. 
[The plane is supposed to be perfectly smooth.] 

1. The angle of the plane is 30°, the weight is 120 kilos: What 
force is required to support the weight (a) acting parallel to the 
plane? (b) acting horizontally? (c) acting at an angle of 60° with 
the plane? 

2. What is the reaction of the plane in the three cases in ex- 
ample 1? 

3. A force of 60 kilos acts parallel to an inclined plane: What 
weight can it support in the following cases — the angle of the 
plane is {a) 20°, (b) 40°. 

4. A weight of 60 kilos rests on an inclined plane: What force 
acting parallel to the plane is required to support it, if the angle 
of the plane has the same values as in example 3? 

5. A force of 50 kilos acts horizontally to an inclined plane : 
What weight can it support if the angle of the plane is 45°? 

6. If a horse can raise 800 kilos vertically, what weight can he 
raise on a railway having a grade of 44 feet to the mile? 



278 ADDITIONAL EXAMPLES. 

XXXII., XXXIII. Wedge— Screw. Articles 233-242; pp. 243-250. 

1. A wedge is isosceles in shape and has an angle of 30° ; if 
P = 20 kilos, what is the resistance on each face? 

2. A wedge is isosceles, and the angle 60°; a force of 100 kilos 
acts at the back: What are the other two forces? 

3. A weight is isosceles, the power acting on the back is 20 
kilos, and the forces on the other sides are 60 kilos each: What 
is the angle of the wedge ? 

4. The wedge is right-angled and the three sides have lengths 
of 25, 20, 15 (back): If P = 100, what are the other two forces? 



5. What weight can be raised by a power of 25 kilos acting on 
a lever-arm f meter long, if the screw has 2 threads to the centi- 
meter? 

6. If a power of 40 kilos acting on an arm 1 meter long can 
support a weight of 8000 kilos, what will be the distance between 
the threads of the screw? 

7. A screw has 1 thread to the centimeter; the circumference 
described by the power is 2 meters : What power is needed to 
support a weight of 3000 kilos? 

8. While the point of application of the power makes a revo- 
lution of 1 meter, the screw advances 6 millimeters; the power is 
50 kilos: What is the weight raised? 

XXXIV. Pendulum. Articles 244-247; pp. 252-261. 

[The length of the seconds pendulum at New York is about 
.9932 meter.] 

1. Required the length of a pendulum at New York which 
will vibrate (a) in ^ second, (b) in 3 seconds, (c) in H seconds. 

2. Required the length of the seconds pendulum where the 
acceleration of gravity is 9.83 (that is, near the pole). 

3. Required the length of a pendulum to vibrate in 2 seconds, 
where the value of g is 9.78. 

4. How many beats in a minute would a pendulum li meters 
long make in New York? 



ANSWERS TO EXAMPLES. 



Pages 12, 13. I. Uniform Motion of Translation or Botation. 
Articles 17-21. 

(1)490.91 miles. (2) 44 feet per second. (3a) 5£f feet per 
second; (35) 117^ yards per minute. (4a) 3 miles; (45) 25 seconds. 
(5) 1 mile. (6) 7.39 miles. (7) 19.01 miles. 

(8) 1047.2 miles per hour, or 1535.9 feet per second. (9) 523.6 
miles per hour, or 767.9 feet per second. (11) .000073. (12a) 
f 7T = 4.189; (125) 12.57 feet per second. (13a) 628.3; (135) 8567.9 
miles. (14a) 5.236 feet per second; (145) 10.472; (14c) 26.18. 

Pages 21, 22. II. Uniformly Accelerated Motion. Articles 22-28. 
A. Falling Bodies. 

(la) 480 ft. per sec. ; (15) 3600 ft. ; (lc) 464 ft. ; (Id) 1296 ft. 
(2a) 18 sec. ; (25) 576 ft. per sec. (3a) 16 sec. ; (35) 4096 ft. (4a) 
11 sec. ; (45) 1936 ft. (5a) 12 sec. ; (55) 2304 ft. (6) i : \ : 1 : 3 : \\ 
the velocities are 8, 16, 32, 96, 144 ft. per sec. (7) T \ : 1 : 1 : 9 : *£; 
the distances are 1 ft., 4, 16, 144, 324 feet. (8a) 31.46 sec; (85) 
1006.86 ft. (9a) 55±- ft.; (95) 108f. (10) 100 ft. (11a) 192 ft., 
256 ft, 384 ft.; (115) when B has fallen 5* sec. (12) 3.94 sec. 
(13) 576 ft. 

B. General Case. — Acceleration =f. 

(1) 8 ft.-per-sec. per sec. (2a) 20; (25) 640 ft.; (2c) 160 ft. per 
sec. (3a) 72 ft. per sec. ; (35) 216 ft. (4) 8. (5a) 249.6 ft. per sec. ; 
(55) 374.4 ft. (6) 7£ sec. (7) 6 sec. and 96 ft. (8) For the earth 
v = 32, 64, 96; and s = 16, 48, 80. The corresponding values for 
the sun are 28 times greater. (9) Yes. (10) Three times as far, 
i.e., 150 ft. 



280 ANSWEKS TO EXAMPLES. 

Pages 30, 31. III. Composition of Velocities. Articles 29-37. 

(la) 1 mile per hour; (16) 9 miles; (lc) 4 miles; (la*) 12 miles. 
(2a) At an angle of 38° 40' with the line drawn directly across 
{BAG, Fig. 11); (26) 6.4 miles per hour; (2c) 24 minutes; (2d) If 
miles down stream (BC, Fig. 11). (3a) 53° 8' up stream (BAC, 
Fig. 12) ; (3b) 3 miles per hour ; (3c) 40 minutes. (4a) At an angle 
of 73° 51' up stream made with the line drawn directly across; 
(46) 1.606 miles per hour; (4c) 1 hr. 26.3 min. (5a) 13° 51' up 
stream measured from the line directly across; (5b) 5.61 miles per 
hour; (5c) 24.7 min. (6) Three fourths of a mile down stream. 
(7a) 13 ft. per sec. ; (76) N. 22° 37' E. (8a) N. 22° 30' E. ; (8b) 
14.78 ft. per sec. (9a) 15 ft. per sec. ; (96) 30° 52' with his own 
direction. (10)0. (11) 2.07 miles per hour due west. (12a) 128.06 
yds. per minute at an angle 38° 40' down stream; (126) 960 yards 
below the starting-point (BC, Fig. 11); (12c) 53° 8' up stream 
(BAC, Fig. 12); (12d) 20 minutes. 

Pages 31, 32. IV. Resolution of Constant Velocities. Article 38. 

(1) 7.79 ft. per sec. (2) 6, 10.39, 12, 10.39, 0. (3) 6.4 ft. per 
sec. parallel AC, and 4.8 parallel AB. (4) 5.2 miles per hour 
north, and 3 miles east. (5a) 8.66; (56) 5. (6) 100 yds. per min. 
and at an angle of 53° 8' up-stream. (7) 11.05 ft. per sec, 
a = 33° 34'. 

Pages 33, 34. Y. Falling down an Inclined Plane. Article 40. 

(la) 16 ft.-per-sec. per sec. ; (16) 128 ft. ; (lc) 64 ft. per sec. ; 
(Id) 56 ft. (2a) 10 seconds; (26) 80 ft. per sec. (3) 3.57 sec. 
(4a) 2 ft.-per-sec. per sec. ; (46) 36 feet. (5a) 4 ft.-per-sec. per sec. ; 
(56) 784 feet. (6a) 1024 feet; (66) 128 ft. per sec. (7) The times 
are 5, 7£, 10, 15, 20 sec. ; the acquired velocity is 160 ft. per sec. 
in all. 

Pages 39, 40. VI. Bodies projected 'vertically downward. Articles 

41, 42. 

(la) 260 ft. per sec. ; (16) 1036 feet; (lc) 244 feet. (2a) 5* sec; 
(26) 196 ft. per sec. (3) 42 ft. per sec. (4) 75 ft. per sec. (5a) 
55 ft. per sec; (56) 675 feet. (6) 21 ft. per sec. (7a) 64 feet ; 
(76) 3| sec. (8a) 1216 feet; (86) 400 feet. ' 



ANSWEES TO EXAMPLES. 281 

Pages 40, 41. VII. Bodies projected vertically upward. Article 44. 

(la, 15) 9 sec; (lc) 1296 feet; (Id) 272 feet and 16 feet. (2a) 3 

or 9 sec. ; (25) 15 or — 3 sec. (3a) 4| sec. ; (35) 8* sec. ; (3c) 15 sec. 

(4) 240 ft. per sec. (5) 12 sec. (6) 192 ft. per sec. (la) 112 ft. 
per sec. ; (7b) same as (7a). (8) 256 ft. per sec. (9) 4 feet from 
the top after i second. (10a) 440 feet; (105) 88 feet; (10c) 88 feet. 
(11) Two seconds after the second ball started. (13) It will 
actually ascend 3 seconds (though apparently falling as seen from 
the balloon) and then descend, reaching the ground after 7 
seconds longer, or 10 seconds in all; the total distance, up and 
down, is 928 feet. 

Pages 41, 42. VIII. Projected up or down a smooth Inclined ^lane. 
Article 45. 
(la) 8 sec. ; (15) 89 feet, per sec. (2a) 96 ft. per sec. ; (25) 12 
sec. (3a) 109 ft. per sec. ; (35) 308 feet; (3c) 101 feet. (4a) 5 sec. ; 
(45) 200 feet; (4c) 48 ft. per sec. ; (U) - 48. 

Page 42. IX. Bodies projected against Friction. Articles 41, 42. 
(la) 12 ft.-per-sec. per sec. ; (15) after 10 sec. ; (lc) 600 ft. (2a) 
5 sec. ; (25) 100 feet; (2c) 16 ft. per sec. (3) 22 seconds and 242 
feet. (4a) 60 miles per hour; (45) four times as far. (5) 4000 
feet; 20 seconds. 

Page 50. X. Projectiles. Articles 47-51. 
(la) 5 seconds; (15) 692.8 feet; (lc) 100 feet. (2) 2 or 3 seconds. 
(3) 7° 14' or 82° 46'. (4) 367.65 ft. per sec. (5) 2$ seconds at a 
distance of 3000 feet. (6a) 1056 ft. per sec. ; (65) 400 feet. (7a) 
120 feet; (75) 48 and 80 ft. per sec. (8c) after 1 second; (Sd) 66 
feet in a horizontal line from the point where it was dropped. 

Page 55. XI. Mass — Density — Volume. Article 56. 
(1) 1 : 4^ (2) 1£ : 1. (3) 9:8. (4) 2f : 1. (5) .49 lb. (6) 1.21 : 1. 
(7)1.1. (8)1.024 

Page 63. XTI. Force of Gravity. Articles 63-65. 
(1) 4 V2 times the earth's radius. (2) g' (sun) : g (earth) = 

—^ : 1; that is, g' = 27.9 g. (3) About | of g. (4) 317 : 1. 

(5) 32.047 (only the difference in distance from the earth's centre 
is considered). 



282 ANSWERS TO EXAMPLES. 

Page 72. XIII. Collision of Inelastic Bodies. Article 70. 

(1) 13£ ft. per sec. (2) 8 ft. per sec. (3) 20 ft. per sec. (4) Re- 
spectively 8, 6, 4|. (5) 4:1. (6) 8 : 7. (7) 7 : 5. (8) :9f ft. per 
sec. (9) 1631.76 ft, per sec. 

Pages 80-82. XIV. General Dynamical Problems. Articles 68 
and 73-76. 

(la) 4 ft.-per-sec. per sec. ; (15) 8 feet. (2a) At a distance equal 
to the earth's radius; (26) one fourth as great as at the surface. 
(3) 5£ ft.-per-sec. per second; the tension (T) = 10 lbs. (4) -^ of 
g. (5a) 100 lbs. ; (5b) 112.5. (6) 20 seconds. (7) 9f feet. (8) 1^ 
oz. (9) T 8 T lb. (10a) 27 lbs.; (10b) 21 lbs. (11) If the velocity is 
gained in 1 second, 40 lbs. (12a) lOf ft.-per-sec. per second; 
(126) 85£ feet. (13) /= 20; 15 oz. = 30 poundals. (14) lf^ lbs. 
(15) 13f sec. (16) 300 lbs. (17) 96 lbs. 

(18a) 8 sec; (186) 256 feet. (19) 5 lbs. (20a) 666f feet. ; (206) 
81 sec. (21a) 215.4 ft. per sec. ; (216) 9.28 sec. 

Pages 88, 89. XV. Centripetal and Centrifugal Forces. Arti- 
cles 77-81. 

(1) /= 112; tension = 70 lbs. (2a) They are increased 4 times; 
(26) they are diminished one half. (3) 986.97 lbs. (4) 16 ft. per 
sec. (5) 2. (6) .726 ton. (7) 74.02 lbs. (8) 16.2 feet (only the 
tension caused by the circular motion is considered). (9) 1 foot. 

Pages 99, 100. XVI. Friction. Articles 82-94 

(1) m - .25. (2) 4.8 lbs. (3) 28 lbs. (4a) // = .364, F= 5.13 
lbs.; (46) n - .364, F= 10.26. (5) 4.657 lbs. (6c) 9.6 and 12 lbs. 
(7) fx = .54. (8a) 6 lbs. ; (86) 5.638 lbs. (9a) P = W sin a - F = 
8.8 lbs.; (96) P = W sin a + F = 15.2 lbs. (10) 10.35 lbs. 
(11) 13.05 lbs. (12) 208.46. (13) 167.42. (14) 6f lbs. (15) 12 lbs. 

Pages 105, 106. XVII. Work. Articles 95-100. 

(1) 125,028 ft.lbs. (2) 102,492.7 ft. lbs. (3) 1,440,000 ft.lbs. (4) 
12,672,000 ft.lbs. (5) 5,940,000 ft.lbs. (6) 600,000 ft.lbs. (7) 
Against gravity 250,000 ft.lbs., against friction 180,000 ft.lbs.; 
total 430,000 ft.lbs. (8) Against gravity 60,000 ft.lbs., against 



ANSWEKS TO EXAMPLES. 283 

friction 24,000 (horizontal surface), 20,7846 (inclined plane); 
total 104,784.6 ft.lbs. 

Pages 124-126. XVIII. Potential and Kinetic Energy. 
Articles 101-108. 

(1) 1800 ft.lbs. (2) 3,267,000,000 ft.lbs. (3) 9,375,000 ft.lbs. per 
minute, or 284.09 horse-power. (4) 250,000 ft.lbs. per minute, 
or 7.58 horse-power. (5) 6924.46 heat-units, and 0.111° C. (6) 
4,500,000 ft.lbs., 3237.41 heat-units. (7) 625 ft.lbs. (8*) 10,000 
feet; (85) four times as far. (9) Same distance. (10a) 6400 feet; 
(105) 40 sec; (10c) 624 feet; (lOd) 304 ft. per sec. (11a) 5000 feet; 
(115)50 sec; (lie) 18,000 ft.lbs. (12) 60.47 feet. (13)/ (the re- 
tardation on the plane) = |(32); (13a) 2000 feet; (135) 12£ sec. 
(14) 343.16 ft. per sec. (the weight of the body does not enter into 
the problem. (15a) 1360 X IT ft.lbs.; (155)6800 feet. (16) B = 
721b6. (17) 240,000 lbs. (18) 37,500 lbs. 

Pages 142, 143. XIX. Parallelogram of Forces. Articles 126-136. 

(1) 25 lbs., a = 73° 44'. (2) 9.097, a = 30° 59'. (3) Q = 16.16, 
y = 83° 29'. (4) (3 = 64° 39', y = 114° 9', P = 5.94; or /3 = 
16° 21', y = 65° 51', P = 1.85. (5) y = 120°, B = 27.71 lbs. (6) 
12.26 lbs. (7) 35 lbs., inclined 53° 8' to the horizontal. (8) 20.78 
lbs. (9) 15 and 20 lbs. (10) 100 lbs., 16° 16' and 73° 44'. (11)23.43 
oz. (12) 5.72 and 11.44 lbs. (13) 41.57 lbs. 

Pages 150, 151. XX. Besolution of Forces. Articles 137, 138. 

(1) 106.05 lbs. K and the same E. (2) 5 lbs., 5.18, 5.77, 7.07, 10, 
19.30, oo . (3) a = 23.83, 5 = 31.11 lbs. (4) a = 34.64 lbs., 5 = 40 
lbs. (5) Each = 39.16 lbs. (6) 579.60 lbs. (7) 15.59 lbs. and 9 lbs. 

Pages 151, 152. XXI. Besolution of Forces along two rectangular 
axes. Articles 140, 141. 

(1) B= 86.6 lbs., and its direction makes an angle of 150° with 
P. (2) B = 100 lbs., at right angles to P. (3) B = 208.3 lbs., and 
makes an angle of 298° 41' (or - 61° 19') with P. (4) B = 73.2 
lbs., and acts due north. (5) B = 346.4 lbs., and acts S. 54° 44' E. 
(6), the forces are in equilibrium. (7) B = 61.22 lbs., and acts 
S. 77° 30' W. ; the system will be kept in equilibrium by a force of 
61.22 lbs acting K 77° 30' E 



284 ANSWERS TO EXAMPLES. 

Pages 159, 160. XXII. Parallel Forces. Articles 143-149. 

(la) B = 12, AG = 28, BG = 20; (lft) B = 30, AB = 135, 
BG = 54; (lc) B = 6, BG = 40, AG = 16; (la") i? = 8, 4£ = 48, 
BG = 84. (2a) $ = 5, AG = 25, 5C= 15; (25) £ = 9, AG= 27, 
4£ = 42; (2c) Q = — 4, £(? = 40, 4(7 = 16; (2d) # = - 4; 
AS = 24, BG = 72. (3) At A 36 lbs., at 5 12 lbs. (4) W = 56 
lbs., at G 24 lbs. (5) At B 9 lbs., at D 1 lb. (the weight of the 
rod is neglected). (6) A carries 84 lbs. , B 60 lbs. (7) At A 4 lbs. , 
at B and G 8 lbs. (8) At A 12 lbs., at B and C 3 lbs. (9) Placed 
at one end. (10) 3l£ and 76^ lbs. 

Page 167. XXIII. Moments. Articles 151-156. 

(1) 72 ft.lbs. (2) 329.10 ft.lbs. (3) 615.64 ft.lbs. (4a) 72 ft.lbs. 
(46) 55.15 ft.lbs. ; (4c) 36 ft.lbs. 

Pages 180-182. XXIV. Centre of Gravity. Articles 159-171. 

(1) 13J- in. from A. (2) 6 in. from G (3) 15 in. from B. (4) 
1 in. from B. (5) 2 in. from B. (6) 2£ lbs. (7) 4 oz. (8) £ lb. 
(9a) A carries 24 lbs., B 36 lbs. ; (9ft) A should stand 8 feet from 
his end. (10) 1^ T in. from the centre toward A. (11) 6 in. from 
C, on a line making an angle of 53° 8' with BG. (12) 16 in. from 
A, on a line bisecting the angle BAG. (13) On a line bisecting 
the right angle, 3 V2 in. from B. (14a) 3 1^2" in. from the centre 
toward A; (14ft) 4f in. from the centre on a line drawn to the 
middle point of AB; (14c) 2 V2 from the centre toward B. (15a) 
2|- in. from the centre, on the line drawn to the middle point of 
GB; (15ft) 2 V2~in. from the centre toward G; (15c) f V2~from the 
centre toward G. (16) 2f in. from the middle point of BG, on the 
line joining the centres of the parallel sides. (17) 1 in. from the 
centre of the original circle. (18a) 2f in. from the centre of the 
larger circle. (19) 3^ in. from the centre of the base. (20) 8 in. 
from B, on the line BD. (21) 11| i n - from the vertex of the tri- 
angle. (22) On the axis, 4 in. from the centre of the larger 
cylinder. 

Pages 189, 190. XXY. Stability. Articles 172-177. 

(la) 45 lbs. ; (1ft) 90 lbs. ; (lc) 125 lbs. and 250 lbs. (2a) 42.43 
lbs.; (2ft) 43.92 lbs. (3)48° 11'. (4) 80 feet vertically. (5a) 45°; 



ANSWEKS TO EXAMPLES. 285 

(55) 67° 10' (the vertex lying up the plane). (6a) 16 lbs.; (65) 48 
lbs. (7a) P=8 lbs. ; (76) 20 lbs. ; (7c) 65 lbs. (8a) 15,406.6 ft.lbs. ; 
(86) 49,442.7 ft.lbs. ; (8c) 8781.8 ft.lbs. 

Pages 213, 214 XXYI. Lever. Articles 185-190. 

(1) 160 lbs. (2) 20 lbs. (3) 12* lbs. (4) 200 lbs., 100 lbs., 871 
lbs. respectively. (5) 37.12 lbs. (6) 46.58 lbs. (7) 180 lbs. (8) 
67.12 vertical (5), and 163.71 lbs. in a direction making an angle of 
12° 13' with a vertical line through F(l). (9) 29.15 lbs. (10) 38.18 
lbs. (11) 96 and 48 lbs. (12) i of the length from the centre 
toward the end having the heavier weight. (13) 8.24 feet from 
the end at which the force 8 acts. 

Page 214. XXVII. Balance. Articles 191-193. 
(1) 14.07 lbs. (2) .837 : 1. (3) 14.06 oz. 

Page 215. XXVIII. Steelyard. Articles 194-196. 

(la) The zero is f inch from C( = GD, Fig. 139); (lb) f in. 
from G( = GG, Fig. 139); (lc) the graduation is to 12ths of an 
inch; (Id) 1 lb. and 20 lbs. (2a) £ in. from G (= CD, Fig. 138); 
(2b) U in. for 1 lb. ; (2c) 15£ lbs. (3) i in. and ^ in. (4) f in. 
from the fulcrum. (5) 6f from the end on which hangs the 
weight. (6a) 18 in. from the end A (Fig. 141); (60) 14f in. fvomB. 

Page 220. XXIX. Wheel and Axle. Articles 202-208. 

(1) W = 1200 lbs. (2) 21,600 lbs. (3) 1920 lbs. (4) P = 88.54 
lbs. (5) 8 feet 4 in. (6) P = 10.61. 

Pages 234, 235. XXX. Pulley. Articles 217-226 . 

(la) 50 lbs. (lb) 150 lbs. (the weight of the platform is neglected 

in each case). (2) n = 4. (3) n = 4. (4) n - 4. (5) TF=~193.18 

(30°), = 173.2 (60°), = 100 (120°), = 51.76 (150°), = (180°). (6) 

P= 31.25, and 31.72 lbs. (7) P= 125 and 125.25. (8) P = 33. 33 and 

32. 97. (9) If w' is the weight of the pulley A, and w" that of B, then 

4 P-\- w" = W + w'. (10) Let the weights of the movable pulleys 

W w' 
D, G, B, A be respectively w', w", w'", w iy , then P — — - -f- — - 4- 

81 81 



286 ANSWEKS TO EXAMPLES. 



w , w , w 



. , 7 + -q- + -o - . If the weights of the pulleys (w) are equal, then 
p W w ( . 1 \ . T^T w (. 1 



Pages 241-243. XXXI. Inclined Plane. Articles 227-232. 

(la) P = 41.04; (15) P= 43.68; (lc) P= 47.39. (2a) i2 = 112.76; 
(26) R = 127.7; (2c) i? = 89.07. (3a) 575.9; (36) 200; (3c) 141.4; 
(M) 115.5; (3c) 101.5; (3/)100. (4a) 17.36; (46) 50; (4c) 70.7; (4d) 
86.6; (4c) 98.5; (4/)100. (5a) 567.1; (5b) 173.2; (5c) 100; (5d) 57.7; 
(5c) 17.6; (5/) 0. (6a) 17.6; (66) 57.7; (6c) 100; (6a) 173.2; (6c)567.1; 
(6/) oo. (7)^ of a ton. (8) 35 and 21 lbs. (9)11,464. lbs. (10) 
■ths of the weight. (11) a = 36° 52', W = 66*. (12) 5*. (13) 
12 feet and 18 feet. 

Page 232. XXXII. Wedge. Articles 233-235. 

(1) 115.2. (2) 70.7. j3) 38° 56'. (4) They are equal. (5) 133* 
and 166*. (6) 1 : 2 : Vd. 

Page 251. XXXIII. Screw. Articles 236-242. 

(1) 9047.8 lbs. (2) .785 in. (3) 12* lbs. (4) 7200 lbs. (5) 
1 : 113.4. (6) 2. (7) £ in. (8) 81,430.3 lbs. 

Page 261. XXXIV. Pendulum. Articles 244-247. 

(la) 9.78 in.; (16) 13.03 feet; (lc) 20.37 feet. (2) 39.21 in. (3) 
13.01 feet. (4) About 91.2 feet. (5) 38.3 times. (6) 37.5 in. 



ANSWERS TO ADDITIONAL EXAMPLES 

(on pages 263-278) 
INTRODUCING THE METRIC UNITS. 



I. 

(1) 864 kilometers. (2) 13f. (3a) 16|; (35) 100. (4a) 5.44 kilo- 
meters; (46) 8ff- minutes. (5) 2.56 kilometers. (6a) 88 (per 
second); (65) 3168 kilometers. 

II., A. 

(1) The distances are 4.9, 19.6, 44.1, 78.4 meters; the velocities 
are 9.8, 19.6, 29.4, 39.2 meters per second. (2a) 98 meters per 
second; (25) 490 meters; (2c) 93.1 meters. (3a) 4f seconds. (35) 
42 meters per second. (4a) 7£ minutes; (45) 275.6 meters. (5a) 5 
seconds; (55) 122.5 seconds. 

II., B. 

(1) 8 meters-per-second per second. (2a) 20 meters-per-second 
per second; (25) .36 kilometers; (2c) 120 meters per second. (3a) 
60 meters per second; (35) 150 meters. (4) Yes. (5) 260 meters. 
(6) 20 meters-per-second per second. 

III. 

(la) 2 kilometers per hour; (15) 18; (lc, Id) 8 and 24. (2a) 
38° 40' downstream; (25) 128.1 meters per minute ; (2c) 20 minutes. 
(3a) 53° 8' up stream; (35) 60 meters per minute; (3c) 33£ minutes. 
(4a) 13 meters per second; (45) K 22° 37' E. (5a) N 67° 30' E.; 
(55) 6.12 meters per second. (6a) 5 meters per second; (65) 36° 52' 
•with his direction. 



288 ANSWERS TO ADDITIONAL EXAMPLES. 

IV. 

(1) 3 meters per second. (2) 4£ and 3f meters per second. (3) 
4.33 and 2.5. (4) 12.99 and 7.5. 

V. 

(la) 4.9 meters-per-second per second; (lb) 39.2; (lc) 19.6. (2a) 
8 seconds; (2b) 19.6 meters per second. (3) 11.89 seconds. (4a) 
1.4 meters-per-second per second; (4&) 90 meters. 

VI. 

(la) 80.6 meters per second; (lb) 324.1 meters. (2a) 5 seconds; 
(2b) 68.1 meters per second. (3) 18.63 meters per second. (4a) 
19.6 meters; (4b) 3f seconds. 

VII. 

(la) and (15) 5 seconds; (lc) 122.5 meters; (Id) 44.1 and 4.9 
meters. (2) 1 or 7f seconds. (3) 56 meters per second. (4) 1\ 
seconds. (5a) 28 meters per second; (5b) after If seconds longer. 

VIII. 
(la) 12 seconds ; (lb) 49.7 meters per second. (2a) 29.6 meters 
per second ; (2b) 79.2 meters. (3a) 56 meters per second ; (36) llf 
seconds. 

IX. 

(la) / = 4 meters-per-second per second ; (lb, lc) after sliding 10 
seconds, and 200 meters. (2a) 20 seconds ; (2b) 120 meters ; (2c) 
10.2 meters per second. (3a) 28.8 kil. per hour ; (Bb) 480 meters. 

X. 

(la) t = 25 seconds; (lb) 5.304 kilometers. (2) 24° 18' or 65° 42'. 
(3) u = 280 meters per second. (4) After 3 seconds, at a distance 
of 1.2 kilometers. (5a) 200 meters per second; (5b) 122.5 meters. 

XI. 
(l)2f:l, (2)2.73:1. (3)773.2. (4) 13.394 kilograms. 

XII. 
(1) 4 times the earth's radius. (2) 277.3 meters-per-second per 



ANSWEES TO ADDITIONAL EXAMPLES. 289 

second. (3) About 1.6 meters-per-second per second. (4) 2.7 
millimeters-per-second per second. 

XIII. 

(1) 5 meters per second. (2) 3 meters per second. (3) 10 meters 
per second. (4) 4, 3, and 2.4 meters per second. (5) 420.84 meters 
per second. 

XIV. 

(la) 3.8 meters-per-second per second ; (lb) 7.6 meters. (2a) At 
a height equal to If the radius of the earth; (2b) 2.04 kilos. (3) 
/= 2.8; tension, 4.286 kilos (4) 20.4 seconds. (5) 28f kilos. 
(6a) 400 grams; (65) 392,000 dynes. (7) 1.4 grams. (8) 1.02 milli- 
grams. 

XV. 

(1) 25 kilos; / = 24.5. (2) 8297^ kilos, 

XVL 

(1) ju = $. (2) 4.8 kilos. (3) 160 grams. (4a) /i = .36, F= 5.07 
kilos. (46) n = .36, F= 10.14 ? (5) 6.235 kilos. 

XVII. 

(1) 1 kilogram-meter = 7.23 ft.lbs. (2) 62,514 kilogram-meters. 
(3) 675,000 kilogram-meters. (4) 100,000 kilogram-meters. (5) 
172,000 kilogram-meters. 

XVIIL 

(1) About 424 kilogram-meters. (2) 200 kilogram-meters. (3) 
50,000,000 kilogram-meters. (4)400,000 kilogram-meters. (5)270 
kilogram-meters. (6) 4 kilometers and 57| seconds. (7) 50 kilos. 
(8) 150,000 kilos. 

XIX 

(1) E= 250 grams; (a) 73° 44'. (2) 11.14 (3) R = 20.78, y = 
120°. (4) 6.13 kilos. (5) 910 grams. (6) 8 kilos. 

XX, XXI. 

(1) 56.4 K 20.52 E. (2) 6.93 and 13.86. (3) 25.21 kilos. (4) 
295.44 kilos. (5) 55424 and 320 grams. 



290 ANSWERS TO ADDITIONAL EXAMPLES. 

(1) R — 303.5, the angle between P and B is - 10° 46'. (2) 100 
kilos in the same direction as P. 

XXII. 

(1) 16 kilos at A, and 8 at B. (2) 10£ at C ; W = 24£. (3) 
A carries 48 and B 32 kilos. (4) On A 2 kilos, and on B and G 
each 4 kilos. (5) 2 centimeters from one end. 

XXIII. 

(1) 72 kilogram-meters. (2) 30 kilogram-meters: the same as 
before. (3) 346.4 kilogram-meters. (4) 25, 23.5, and 8.55 kilo- 
gram-meters. 

XXIY., XXY. 

(1) 21 centimeters from A. (2) 25 millimeters from B. (3) 270 
grams. (4) 10.02 gr. (5) 125 from the centre toward A. (6) 69 mil- 
limeters from A, on the line drawn to the middle point of BC. 
(la) 21 kilos; (76) 56 kilos. (8a) 7.54 kilos ; (85) 38.63 kilos ; (8c) 
6.63 kilos. (9a) 811.4 kilogram-meters ; {9b) 3090.2 ; (9c) 450.85. 

XXYI. 

(1) 53£ kilos. (2) 20 kilos. (3) 3 kilos. (4) 24.24 kilos. (5) 
69.28 kilos. (6) 15.81 kilos. (7) 21.21 kilos. (8) 1 meter 856.4 mm. 
from the force 8. 

XXVII., XXVIII. 

(1) 6 kilos 245 grams. (2) 1 : 1.08. (3) 1041.7 milligrams. (4a) 
12 millimeters from C; (45) 80 millimeters; (4c) 7.1 kilos. (5) 25 
millimeters from the fulcrum. (6) 90 millimeters from the end 
on which hangs the weight. 

XXIX, XXX. 

(1) 900 kilos. (2) 333£ kilos. (3) 50.4 kilos. (4) 2£ meters. 
-(5) 4 pulleys. (6) 6 pulleys. (7) 5 pulleys. (8) 54.12 (2 a = 
45°), 130.66 (2 a = 135°). (9) 50 kilos, 50.25 kilos. 

XXXI. 

(la) 60 kilos; (15) 69.28 kilos; (lc) 120 kilos. (2a) 103.92; (2fc) 



ANSWEKS TO ADDITIONAL EXAMPLES. 291 

138.56; (2c) 0. (3a) 175.4; (36) 93.34. (4a) 20.52 ; (45) 38.57. (5)50 
kilos. (6) 96,000 kilos. 

XXXII., XXXIII. 

(1) 38.64. (2) Each 100 kilos. (3) 19° 12'. (4) 100, 133*, 166*.— 
(5) 25,132.7 kilos. (6) 31.42 millimeters. (7) 15 kilos. (8)8333* 
kilos. 

XXXIV. 

(la) .110 meter; (lb) 8.939 meters; (lc) 2.235 meters. (2) .9962 
meters. (3) 3.965 meters. (4) 48.8. 



